Mathematics PREVIOUS YEAR QUESTIONS OF Angles and Trigonometric Ratios For NDA

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PREVIOUS YEAR QUESTIONS OF Angles and Trigonometric Ratios For NDA

Previous Year Questions

Set-1

Q 2200891718

On simplifying ` ( sin ^3 A +sin 3A)/(sin A) + (cos ^3 A - cos 3A)/(cos A)`
, we get

NDA Paper 1 2015

(A)

`sin 3A`

(B)

`cos 3A`

(C)

`sin A+ cos A`

(D)

`3`

Solution:

` ( sin ^3 A +sin 3A)/(sin A) + (cos ^3 A - cos 3A)/(cos A)`

`= ( sin ^3 A + (3sin A - 4 sin ^3 A))/(sin A) + (cos ^3 A - (4 cos ^3 A - 3 cos A))/(cos A)`

`[∵ sin 3 theta = 3 sin theta - 4sin ^3 theta , cos 3 theta = 4 cos^3 theta - 3cos theta]`

`= (- 3 sin^ 2 A + 3) + (- 3 cos^ 2 A + 3)`

`= 6- 3(sin^2 A+ cos^2 A) = 6- 3 (1)= 3`

Correct Answer is `=>` (D) `3`

Q 1772345236

What is value of ` ( cos 224^0 - cot 134^0)/( cot 226^0 + cot 316^0)` ?
NDA Paper 1 2014

(A)

`- cosec quad 88^0`

(B)

`- cosec quad 2^0`

(C)

`- cosec quad 44^0`

(D)

`- cosec quad 46^0`

Solution:

` ( cot 224^0 - cot 134^0)/( cot 226^0 + cot 316^0)`

` = (cot (360^0 - 136^0) -cot (90^0 + 44^0))/( cot (360^0 -134) + cot (360^0 - 44^0))`

` [ ∵cot (360^0 - theta ) =-cot theta ` and `cot (90^0 + theta) = - tan theta ]`

` = ( - cot 136^0 + tan 44^0)/(-cot 134^0 -cot 44^0)`

` = ( -cot (90^0 + 46^0) + tan 44^0) / ( -cot (90^0 + 44^0) - cot 44^0)`

` = ( tan 46^0 + tan 44^0)/(tan 44^0 - cot (90^0 - 46^0 ) )`

` [ ∵ cot (90^0 + theta ) = -tan theta]`

` = ( (sin 46^0)/(cos 46^0) + ( sin 44^0)/(cos 44^0) )/( (sin 44^0)/(cos 44^0) + ( sin 46^0)/(cos 46^0) )`

` = ( sin 46^0 . cos 44^0 + sin 44^0 . cos 46^0)/( sin 44^0 . cos 46^0 - sin 46^0 . cos 44^0)`

` = ( sin (46^0 + 44^0))/(sin (44^0 - 46^0 ) )`

` [ ∵ sin (A+ B) =sin A. cos B +cos A. sin B,`

`sin (A- B)= sin A. cos B - cos A. sin B]`

` = ( sin 90)/(sin (-2))`

` = 1/ ( - sin 2)`

` [ ∵sin 90^0 = 1, sin (- theta ) = - sin theta ` and ` 1/ (sin theta) = =cosec theta ]`

` = - cosec quad 2`

Correct Answer is `=>` (B) `- cosec quad 2^0`

Q 1762545435

What is `cos 20^0 +cos 100^0 +cos 140^0` equal to?
NDA Paper 1 2014

(A)

`2`

(B)

`1`

(C)

`1/2`

(D)

`0`

Solution:

We have, `cos 20^0 + cos100^0 + cos 140^0`

` = (cos 140^0 + cos 20^0) +cos 100^0`

` = 2 cos ( (140^2 + 20^2)/2) . cos ((140^0 -20^0)/2) + cos 100^0`

` = 2 cos ((160^0)/2) . cos ((120^0)/2) + cos 100^0`

` = 2 cos (80^0) . cos 60^0 + cos 100^0`

` = 2.cos 80^0 (1/2) + cos 100^0`

`= cos 80^0 + cos 100^0`

` = 2 cos ( ( 80^0 + 100^0)/2) . cos ((100 - 80^0)/2)`

` = 2 cos ((180^0)/2) . cos ((20^0)/2)`

` = 2cos 90^0 . cos 10^0`

` ( ∵ cos C + cos D = 2 cos ((C + D)/2) . cos ((C - D)/2 ))`

` = 2 xx 0 xx cos 10^0`

` = 0`

Correct Answer is `=>` (D) `0`

Q 1712778630

What is the value of `cos 36^0`?
NDA Paper 1 2014

(A)

` (sqrt(5) - 1) /4 `

(B)

` (sqrt(5) + 1) /4 `

(C)

` sqrt(10 + 2sqrt(5))/4`

(D)

` sqrt(10 - 2sqrt(5))/4`

Solution:

We take, `5 theta = 180^0`

`=> 3theta +2 theta =100^0`

` => 2theta = 180^0 -3theta `

On taking cos both sides, we get

`=> cos 2theta =cos (180^0 - 3theta )`

` => cos 2theta = - cos 3theta `

` => 2 cos^2 theta - 1 = - ( 4 cos^3 theta - 3 cos theta)`

` => 2 cos^2 theta - 1 =- 4 cos^3 theta + 3 cos theta`

` => 4 cos^3 theta + 2 cos^2 theta - 3 cos theta - 1 = 0`

` => 4 cos^2 theta ( cos theta + 1) - 2 cos theta ( cos theta + 1) -1 ( cos theta + 1) = 0`

` => ( cos theta + 1) ( 4 cos^2 theta - 2 cos theta -1 ) = 0`

` => cos theta + 1 = 0 ` or ` 4 cos^2 theta - 2 cos theta -1 = 0`

` => cos theta = -1`

` => cos theta = ( 2 pm sqrt(4 - 4 xx (4) (-1)))/( 2 xx 4)`

` => cos theta = ( 2 pm sqrt( 4 + 16))/8 = ( 2pm sqrt(20))/8`

` => cos theta = ( 2 pm 2 sqrt(5))/8`

` => cos theta = ( 1 pm sqrt(5))/4`

Put `theta = 36^0`, we get

` cos 36^0 = (sqrt(5) + 1) /4 `

Correct Answer is `=>` (B) ` (sqrt(5) + 1) /4 `

Q 1780101917

If `cot A = 2` and `cotB = 3`, then what is the value of
`cot(A +B)?`
NDA Paper 1 2014

(A)

` pi/6`

(B)

`pi`

(C)

`pi/2`

(D)

`pi/4`

Solution:

We have, `cot A= 2` and `cot B = 3`

To find `A+ B`

Consider, `cot (A + B) = (cot A cot B - 1)/(cot A + cot B) = (6 - 1)/(2 +3) = 5/5 = 1`

`=> cot (A + B) =cot ( pi/4) => A + B = pi/4`

Correct Answer is `=>` (D) `pi/4`

Q 1730445312

What is `sin^2 66 1^0/2 - sin^2 23 1^0/2` equal to
NDA Paper 1 2014

(A)

`sin 47^0`

(B)

`cos 47^0`

(C)

`2 sin 47^0`

(D)

`2 cos 47^0`

Solution:

Consider,

`sin^2 66 1^0/2 - sin^2 23 1^0/2`

` = [ sin ( 90^0 - 23 1^0/2 )]^2 - sin^2 23 1^0/2`

` = cos^2 23 1^0/2 - sin^2 23 1^0/2`

` = cos 2 ( 23 1^0/2) quad ( ∵ cos 2A = cos^2 A - sin^2 A)`

` = cos [ 2 xx ( (47)/2 )^0 ]= cos 47^0`

Correct Answer is `=>` (B) `cos 47^0`

Q 1780256117

In a `Delta ABC`, if `sin A -cos B = cos C`, then what is `B` equal to?
NDA Paper 1 2014

(A)

`pi`

(B)

`pi/3`

(C)

`pi/2`

(D)

`pi/4`

Solution:

In a `Delta ABC`, we have

`=> sin A - cos B = cos C => sin A = cos B + cos C`

`=> 2 sin (A/2) . cos .A/2 =2 cos ((B +C)/2) . cos ((B - C)/2)`

`[∵ sin2A = 2sin A . cos A`

and `cos B +cos C = 2 cos ((B +C)/2) . cos ((B - C)/2) ]`

`=> 2 sin (A/2) . cos (A/2) =2 cos ( 90^0 - A/2 ) . cos ((B - C)/2) `

` [ ∵ A+B+C=180^0 => ((B - C)/2) = 90^0 - A/2 ]`

`=> 2 sin (A/2) . cos (A/2) = 2. sin (A/2) . cos ((B - C)/2) `

` => cos (A/2) = cos ((B - C)/2) `

`=> A/2 = (B-C)/2 `

`=> A + C = B ` .........(1)

also `A + C =180^0 - B` .........(2)

` 180^0 - B = B`

` => 2B = 180^0`

`:. B = 90^0`

Correct Answer is `=>` (C) `pi/2`

Q 1730456312

If `(sin (x + y))/(sin (x - y) ) = (a + b)/( a - b)`, then what is `(tan x)/(tan y)` equal to?

NDA Paper 1 2014

(A)

`b/a`

(B)

`a/b`

(C)

`ab`

(D)

`1`

Solution:

`(sin (x + y))/(sin (x - y) ) = (a + b)/( a - b)`

Applying componendo and dividendo, we get

` (sin (x + y) +sin (x - y) )/(sin (x + y)- sin (x- y)) = ( (a + b)+( a - b) )/ ( (a + b)-( a - b))`

` => ( 2 sin x. cos y)/(2 cos x . sin y) = (2a)/(2b) => tan x . cot y = a/b`

`:. (tan x)/(tan y) = a/b`

Correct Answer is `=>` (B) `a/b`

Q 1710556419

If `sin A sin (60^0 - A) sin (60^0 + A) = k sin 3A`, then what
is `k` equal to?
NDA Paper 1 2014

(A)

`1//4`

(B)

`1//2`

(C)

`1`

(D)

`4`

Solution:

We have,

`sin A. sin (60^0 - A) sin (60^0 + A) = k. sin 3A`

`=> sin A . (sin 3A)/( 4 sin A) = K . sin 3A`

`[ ∵ sin (60^0 + A).sin (60^0 -A) = (sin 3A)/( 4 sin A) ]`

` => (sin 3A)/4 = K . sin 3A`

`:. k = 1/4`

Correct Answer is `=>` (A) `1//4`

Q 1770056816

Which one of the following is one of the solutions of the
equation `tan 2 theta . tan theta = 1`?
NDA Paper 1 2014

(A)

` pi/(12)`

(B)

`pi/6`

(C)

`pi/4`

(D)

`pi/3`

Solution:

We have,

`tan 2 theta . tan theta = 1`

`=> (2 tan theta)/(1 - tan^2 theta) . tan theta = 1`

`=> 2 tan^2 theta = 1 - tan^2 theta => 3 tan^2 theta = 1`

`=> tan^2 theta = 1/3 = (1/sqrt(3))^2`

` => tan^2 theta = tan^2 (30^0) = tan^2 (pi/6) => theta = npi pm pi/6`

` :. theta = pi/6`

Correct Answer is `=>` (B) `pi/6`

Q 1700191918

Given, `16 sin^5 x = p sin 5x + q sin 3x + r sin x`

What is the value of `p?`
NDA Paper 1 2014

(A)

`1`

(B)

`2`

(C)

`-1`

(D)

`-2`

Solution:

Consider,

`16 sin^5 x = 16(sin^2 x)^2 -sin x`

` = 16 ((1 - cos 2x) /2)^2 . sin x`

` = 4 (1 + cos^2 2x- 2 cos 2x) . sin x`

` = 4 (1 + ( 1 + cos 4x)/2 - 2 cos 2x) . sin x`

` = 4/3 (3 + cos 4x - 4 cos 2x) . sin x`

` = (6 + 2 cos 4x - 8cos 2x)sin x`

` = 6sin x + 2 sin x cos 4x - 8cos 2x . sin x`

` = 6sin x +sin 5x- sin 3x - 4(sin 3x- sin x)`

` [∵ 2 sin A cos B = sin (A+ B)+ sin (A- B)]`

` = 6sin x + sin 5x - sin 3x - 4sin 3x + 4sin x`

` = sin 5x - 5sin 3x + 10 sin x`

Clearly, `p = 1`, hence option `(a)` is correct.

Correct Answer is `=>` (A) `1`

Q 1710191919

Given, `16 sin^5 x = p sin 5x + q sin 3x + r sin x`

What is value of `q?`

NDA Paper 1 2014

(A)

`3`

(B)

`5`

(C)

`10`

(D)

`-5`

Correct Answer is `=>` (D) `-5`

Q 1711101020

Given, `16 sin^5 x = p sin 5x + q sin 3x + r sin x`

What is the value of `r?`
NDA Paper 1 2014

(A)

`5`

(B)

`8`

(C)

`10`

(D)

`-10`

Correct Answer is `=>` (C) `10`

Q 2309756618

What is the angle (in circular measure) between
the hour hand and the minute hand of a clock
when the time is half past `4` ?
NDA Paper 1 2013

(A)

`60^o`

(B)

`45^o`

(C)

`30^o`

(D)

None of these

Solution:

At `30` min past 4, the minute hand is at 6 and hour hand
slightly advanced from 4.

Since, 10 min small parts between 4 and 6.
:. Angles `= 10 xx 6^o = 60°`

Since, the hour hand slightly moves from 4 O' clock.

So, the angle is lesser than `60°.`

·: In `30` min, minute hand moves `= 30 xx (1^o)/2 = 15°`

:. Required angle `= 60^o-15^o=45^o`

Correct Answer is `=>` (B) `45^o`

Q 2460323215

What is the angle (in circular measure) between
the hour hand and the minute hand of a clock
when the time is half past 4?
NDA Paper 1 2013

(A)

`60^0`

(B)

`45^0`

(C)

`30^0`

(D)

None of these

Solution:

At `30` min past `4`, the minute hand is at `6` and hour hand
slightly advanced from `4`.
Since, `10` min small parts between `4` and `6`.
`therefore` Angles `= 10 xx 6^0 = 60^0`
Since, the hour hand slightly moves from `4` O' clock.
So, the angle is lesser than `60^0`.

`because` in 30 min minute hand moves = `30xx1^0/2 = 15^0`

`therefore` Required angle = `60^0-15^0 = 45^0`

Correct Answer is `=>` (B) `45^0`

Q 2420523411

What is `(cot54^0)/(tan36^0)+(tan20^0)/(cot70^0)` equal to ?
NDA Paper 1 2013

(A)

`0`

(B)

`1`

(C)

`2`

(D)

`3`

Solution:

`(cot54^0)/(tan36^0)+(tan20^0)/(cot70^0) = (cot54^0)/(tan(90^0-54^0))+(tan20^0)/(cot(90^0-20^0))`

` = (cot54^0)/(cot54^0)+(tan20^0)/(tan20^0) = 1+1 = 2`

Correct Answer is `=>` (C) `2`

Q 2440723613

If `cos x = 1/3` then what is `sin x ·cot x ·cosec x · tan x` equal to ?
NDA Paper 1 2013

(A)

`2/3`

(B)

`3/2`

(C)

`2`

(D)

`1`

Solution:

Given that, `cos x = 1/3`

We have. `sin x. cot x · cosec x ·tan x`

` = sinx * (cosx)/(sinx) * 1/(sinx) * (sinx)/(cosx) = 1`

Correct Answer is `=>` (D) `1`

Q 2400023818

Consider the following statements

`I. tan(pi/6)`

`II. tan((3pi)/4)`

`III. tan((5pi)/4)`

`IV. tan((2pi)/3)`

Which one of the following is the correct order?
NDA Paper 1 2013

(A)

`I < IV < II < Ill`

(B)

`IV < II < I < Ill`

(C)

`IV < II < Ill < I`

(D)

`I < IV< Ill < II`

Solution:

`I. tan(pi/6) = tan30^0 = 1/sqrt3`

`II. tan((3pi)/4) = tan135^0 = tan(90^0+45^0) = -cot45^0 = -1`

`III. tan((5pi)/4) = tan225^0 = tan(180^0+45^0) = tan45^0 = 1`

`IV. tan((2pi)/3) = tan120^0 = tan(90^0+30^0) = -cot30^0 = -sqrt3`

So, the correct order is IV< II< I< Ill.

Correct Answer is `=>` (B) `IV < II < I < Ill`

Q 2430134012

If `cosec theta + cot theta = c`, then what is `cos theta` equal to?
NDA Paper 1 2013

(A)

`c/(c^2-1)`

(B)

`c/(c^2+1)`

(C)

`(c^2-1)/(c^2+1)`

(D)

None of these

Solution:

Given that, `cosec theta + cot theta = c`

`=> 1/(sintheta)+(costheta)/(sintheta) = c`

`=> (1+costheta)/(sintheta) = c`

`=> (1+2cos^2 (theta/2) -1)/(2 sin (theta/2) * cos (theta/2)) = c => (cos(theta/2))/(sin(theta/2)) = cot(theta/2) = c`

`=> tan (theta/2) = 1/c`..............(i)

`therefore costheta = (1-tan^2 (theta/2))/(1+tan^2 (theta/2)) = (1-(1/c)^2)/(1+(1/c)^2)` [from eq(i)]

`= (c^2-1)/(c^2+1)`

Correct Answer is `=>` (C) `(c^2-1)/(c^2+1)`

Q 2440334213

If `sin theta + 2cos theta = 1` , then what is `2 sin theta - cos theta`
equal to'?
NDA Paper 1 2013

(A)

`0`

(B)

`1`

(C)

`2`

(D)

`4`

Solution:

Given curve, `sintheta +2costheta = 1`

On squaring both sides, we get

`(sin theta+2 cos theta )^2 = 1`

`=> sin^2 theta+4 cos^2 theta +4 sin theta * cos theta = 1`

`=> (1-cos^2 theta )+4(1-sin^2 theta )+4 sin theta * cos theta = 1` `(because sin^2 theta +cos^2 theta = 1)`

`=> 4sin^2 theta+cos^2 theta-4sintheta * costheta = 4`

`=> (2 sin theta -cos theta)^2 = 4 = (2)^2`

`therefore 2sintheta -costheta = 2`

Correct Answer is `=>` (C) `2`

Q 2470434316

If `A+ B = 90^0`, then what is the value

`sqrt(sinAsecB-sinAcosB?)`
NDA Paper 1 2013

(A)

`sin A`

(B)

`cos A`

(C)

`tanA`

(D)

`0`

Solution:

Given that `A+ B = 90^0` ............(i)

`sqrt(sinAsecB-sinAcosB)`

`sqrt(sinAsec(90^0-A)-sinAcos(90^0-A))`

`sqrt(sinA * cosecA - sinA * sinA)`

` = sqrt(sinA * 1/(sinA) - sin^2A)`

`= sqrt(1-sin^2A)`

`sqrt(cos^2A) = cosA`

Correct Answer is `=>` (B) `cos A`

Q 2440534413

What is `tan^4 A- sec^4 A+ tan^2 A+ sec^2 A` equal
to'?
NDA Paper 1 2013

(A)

`0`

(B)

`1`

(C)

`2`

(D)

`-1`

Solution:

`tan^4 A-sec^4 A+ tan^2 A+ sec^2 A`
`= (tan^2 A)^2 -(sec^2 A)^2+ (tan^2 A +sec^2 A)`

`= (tan^2 A-sec^2 A)(tan^2 A +sec^2 A)+(tan^2 A+sec^2 A)`
`= (-1)(tan^2 A+ sec^2 A)+ (tan^2 A+ sec^2 A) (because sec^2 A- tan^2 A=1)`

`= -(tan^2 A+ sec^2 A)+ (tan^2 A+ sec^2 A)= 0`

Correct Answer is `=>` (A) `0`

Q 2430634512

What is the value of `tan 105^0 ?`
NDA Paper 1 2013

(A)

`(sqrt3+1)/(sqrt3-1)`

(B)

`(sqrt3+1)/(1-sqrt3)`

(C)

`(sqrt3-1)/(sqrt3+1)`

(D)

`(sqrt3+2)/(sqrt3-1)`

Solution:

`tan105^0 = tan(60^0+45^0)`

` = (tan60^0+tan45^0)/(1-tan60^0*tan45^0)` `[because tan(A+B) = (tanA+tanB)/(1-tanA*tanB)]`

` = (sqrt3+1)/(1- sqrt3 *1) `

` = (sqrt3+1)/(1- sqrt3)`

Correct Answer is `=>` (B) `(sqrt3+1)/(1-sqrt3)`

Q 2420734611

If `tan A = x + 1` and `tan B = x - 1`, then
`x^2 tan(A- B)` has the value
NDA Paper 1 2013

(A)

`1`

(B)

`x`

(C)

`0`

(D)

`2`

Solution:

Given that, `tan A = x + 1` ........(i)

and `tanB = x- 1` ... (ii)

Now `x^2tan(A-B) = x^2((tanA-tanB)/(1+tanA *tanB))`

` = x^2{((x+1)-(x-1))/(1+(x+1)*(x-1))}` [from Eqs(ii)]

` = x^2{2/(1+x^2-1)} = x^2 * 2/x^2`

` = 2`

Correct Answer is `=>` (D) `2`

Q 2410834710

·what is the value of `(sin^4 theta - cos^4 theta + 1) cosec^2 theta ?`
NDA Paper 1 2013

(A)

`-2`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

`(sin^4 theta - cos^4 theta + 1) cosec^2 theta `

` = {(sin^2theta-cos^2 theta)(sin^2 theta+cos^2 theta)+1} * cosec^2 theta`

` = {(sin^2 theta -cos^2 theta )}*1+1}cosec^2 theta`

` = {sin^2 theta -cos^2 theta +1} cosec^2 theta`

` = {2sin^2 theta } * 1/(sin^2 theta) = 2`

Correct Answer is `=>` (D) `2`

Q 2410845710

What is the value of `sin 15^0`?
NDA Paper 1 2012

(A)

`(sqrt3-1)/(2sqrt2)`

(B)

`(sqrt3+1)/(2 sqrt2)`

(C)

`(sqrt3-1)/(sqrt3+1)`

(D)

`(sqrt3+1)/(sqrt3-1)`

Solution:

`sin15^0 = sin(45^0-30^0)`

` = sin45^0cos30^0-cos45^0sin30^0`

` = 1/sqrt2 * sqrt3/2 - 1/sqrt2 * 1/2 = (sqrt3-1) /(2sqrt2)`

Correct Answer is `=>` (A) `(sqrt3-1)/(2sqrt2)`

Q 2470845716

If `4 sin^2 theta = 1`, where `0 < e < 2pi` , then how many
values does `theta` take?
NDA Paper 1 2012

(A)

`1`

(B)

`2`

(C)

`4`

(D)

None of these

Solution:

Given, `4sin^2 theta = 1 => sin^2 theta = 1/4`

`therefore sintheta = pm1/2`

Since, `theta` llies in the interval `0 < theta < 2pi` . So, the values of `theta` are

`pi/6 , (5pi)/6 , pi+pi/6` and `2pi - pi/6`

` pi/6 , (5pi)/6 , (7pi)/6` and `(11pi)/6`

Correct Answer is `=>` (C) `4`

Q 2440045813

If `sec alpha = 13/5` where `270^0 < alpha < 360^0` then what is the value of `sinalpha ?`

NDA Paper 1 2012

(A)

`5/13`

(B)

`12/13`

(C)

`-12/13`

(D)

`-13/12`

Solution:

`because secalpha = 13/5`

`therefore cos alpha = 5/13`

Now `sinalpha = sqrt(1-cos^2alpha) = sqrt(1-25/169) = sqrt(144/169) = -12/13`

(since, `sin alpha` is negative in fourth quadrant i.e.`,270 < a < 360^0`)

Correct Answer is `=>` (C) `-12/13`

Q 2450045814

What is the value of `tan(-585^0)?`
NDA Paper 1 2012

(A)

`1`

(B)

`-1`

(C)

`-sqrt2`

(D)

`-sqrt3`

Solution:

`tan(-585^0) = -tan(585^0)`

` = -tan(180xx3+45) = -tan45 = -1`

Correct Answer is `=>` (B) `-1`

Q 2480045817

Consider the following statements
I. The value of `cos 46^0- sin 46^0` is positive.
II. The value of `cos 44^0- sin 44^0` is negative.
Which one of the above statements is/are correct?
NDA Paper 1 2012

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

L `cos 46^0 - sin46^0 = cos 46^0 - cos 44^0`

` = 2 sin45^0 sin(-1) = 2 * 1/sqrt2 (-sin1) = -sqrt2 sin1`

A negative quantity will be obtained.
IL `cos44^0- sin44^0 = cos44^0- cos46^0`
`= 2sin45^0 Sin(1)`

` = 2 * 1/sqrt2 sin1 = sqrt2 sin1`

A positive quantity will be obtained.

Correct Answer is `=>` (D) Neither I nor II

Q 2400045818

The line making an angle `(-120^0)` with X-axis is
situated in the
NDA Paper 1 2012

(A)

first quadrant

(B)

second quadrant

(C)

third quadrant

(D)

fourth quadrant

Solution:

Here, we are given only that the line making an angle
(-120°)with theX-axis but we are not given any point, form which
the line the passes. Drawing for any equation of line we should
know a point and an angle about that line

Q 2410045819

The angle subtended at the centre of a circle of
radius `3 ` cm by an arc of length `1` cm is
NDA Paper 1 2012

(A)

`(30^0)/pi`

(B)

`(60^0)/pi`

(C)

`60^0`

(D)

None of these

Solution:

`because theta = l/r = 1/3 = (1/3)^0xx(180^0)/pi = 60^0/pi`

Correct Answer is `=>` (B) `(60^0)/pi`

Q 2450156014

If `sinA = 2/sqrt5` and `cosB = 1/sqrt(10)` where A and B are acute angles then what is the value of `A+B ?`
NDA Paper 1 2012

(A)

`135^0`

(B)

`90^0`

(C)

`75^0`

(D)

`60^0`

Solution:

Given that, A and B are acute angles.

`A < 90^0` and `B < 90^0` and `sinA = 2/sqrt5 , cos B = 1/sqrt(10)`

We know that `sin^2 theta+cos^2 theta = 1`

`therefore cosA = sqrt(1-sin^2A) = sqrt(1-4/5) = 1/sqrt5`

`sinB = sqrt(1-cos^2B) = sqrt(1-1/10) = 3/sqrt(10)`

`because sin(A+B) = sinA *cosB+cosA*sinB`

` = 2/sqrt5 * 1/sqrt(10)+1/sqrt5*3/sqrt(10)`

` = (2+3)/(sqrt5 sqrt(10)) = 5/(sqrt5 sqrt(10)) = sqrt5/sqrt(10) = 1/sqrt2 = sin135^0`

`therefore A+B = 135^0`

Correct Answer is `=>` (A) `135^0`

Q 2410156019

What is the value of `tan(pi/12)?`
NDA Paper 1 2012

(A)

`2-sqrt3`

(B)

`2+sqrt3`

(C)

`sqrt2-sqrt3`

(D)

`sqrt3-sqrt2`

Solution:

`tan(pi/12) = tan(180^0/12^0) = tan15^0`

`= tan (45^0-30^0) = (tan45^0-tan30^0)/(1+tan45^0*tan30^0)`

` = (1-1/sqrt3)/(1+1/sqrt3) = (sqrt3-1)/(sqrt3+1) * (sqrt3-1)/(sqrt3-1)`

` = (sqrt3-1)^2/(3-1) = (3+1-2sqrt3)/2 = 2-sqrt3`

Correct Answer is `=>` (A) `2-sqrt3`

Q 2430456312

If `cosec theta -cottheta = 1/sqrt3` where `theta ne 0` then what is the value of `cos theta?`
NDA Paper 1 2012

(A)

`0`

(B)

`sqrt3/2`

(C)

`1/2`

(D)

`1/sqrt2`

Solution:

`cosec theta -cottheta = 1/sqrt3` where `theta ne 0`

` => 1/(sintheta ) -(costheta)/(sintheta) = 1/sqrt3`

` => (1-costheta)/(sintheta) = 1/sqrt3`

`=> {(1-(1-2sin^2 (theta/2)))/(sintheta)} = 1/sqrt3`

`=> (2 sin^2 (theta/2))/(2 sin (theta/2) * cos(theta/2)) = 1/sqrt3`

` => tan(theta/2) = tan30^0 => theta/2 = 30^0`

`=>theta = 60^0`

`therefore costheta = cos60^0 = 1/2`

Correct Answer is `=>` (C) `1/2`

Q 2410556410

What is the maximum value of
`sin3 theta cos 2 theta +cos 3 theta sin2 theta?`
NDA Paper 1 2012

(A)

`1`

(B)

`2`

(C)

`4`

(D)

`10`

Solution:

Let `f(theta) = sin3theta.cos2 theta + cos3theta · sin2 theta`

`=sin (3theta + 2theta)= sin 5theta`

`[because sin (A+ B) =sin A·cos B +cos A ·Sin B)`

We know that, `-1 le sin 5theta le 1 => -1 le f(0) le 1`

So, the maximum value of `f(theta)` is 1.

Correct Answer is `=>` (A) `1`

Q 2440656513

What is the value of `sin (1920^0)?`
NDA Paper 1 2012

(A)

`1/2`

(B)

`1/sqrt2`

(C)

`sqrt3/2`

(D)

`1/3`

Solution:

`sin(360^0xx5^0+120^0)` `[because sin(360^0+theta) = sintheta]`

` = sin120^0`

`= sin(90^0+30^0)` `[because sin(90^0+theta) = costheta]`

` = cos30^0 = sqrt3/2`

Correct Answer is `=>` (C) `sqrt3/2`

Q 2400756618

If `tan theta + sec theta = 4`, then what is the value of `sintheta?`
NDA Paper 1 2012

(A)

`8/17`

(B)

`8/15`

(C)

`15/17`

(D)

`23/32`

Solution:

Given that, `tantheta+sectheta = 4`

`=> (sintheta)/(costheta)+1/(costheta) = 4`

`=> (1+sintheta)/(costheta) = 4`

`{(cos^2 theta +sin^2 (theta/2)+2sin (theta/2)* cos(theta/2)}/{cos^2 (theta/2)-sin^2 (theta/2))} = 4`

` => ((cos(theta/2)+sin9theta/2))/((cos(theta/2)+sin(theta/2))(cos(theta/2)-sin(theta/2))) = 4`

`=> (cos(theta/2)+sin(theta/2))^2/(cos(theta/2)-sin(theta/2)) = 4`

`=> (1+tan(theta/2))/(1-tan(theta/2)) = 4`

`=> 1+tan(theta/2) = 4-4tan(theta/2)`

`=> 5tan(theat/2) = 3`.......(i)

`therefore sintheta = (2tan(theta/2))/(1+tan^2(theta/2)) = (2*3/5)/(1+(9/25)`

` = (6/5)/(34/25) = 30/34`

` = 15/17`

Correct Answer is `=>` (C) `15/17`

Q 2480167917

If `cot A · cot B = 2,` then what is the value of
`cos(A +B) · sec (A- B)?`
NDA Paper 1 2012

(A)

`1/3`

(B)

`2/3`

(C)

`1`

(D)

`-1`

Solution:

Given that `cot A · cot B = 2,` ................(i)

`=> (cosA * cosB)/(sinA * sinB) = 2`

`=> (cosA * cos B- sinA * sinB)/(cosA * cosB+ sinA * sinB) = (2-1)/(2+1)`

(by componendo and dividendo rule)

`=> (cos (A+B))/(cos(A-B)) = 1/3`

`therefore cos(A+B) * sec(A-B) = 1/3`

Correct Answer is `=>` (A) `1/3`

Q 2430178012

Let `sin (A+ B) = 1` and `sin(A-B) = 1/2` where `AB in [0, pi/2]`

What is the value ofA'?
NDA Paper 1 2012

(A)

`pi/6`

(B)

`pi/3`

(C)

`pi/4`

(D)

`pi/8`

Solution:

`sin (A+ B) = 1` and `sin(A-B) = 1/2` where `AB in i[0, pi/2]`

`=> sin(A+B) = sin(pi/2)`

`=> A+B = pi/2` ...............(i)

and `sin(A-B) = 1/2 => sin(A-B) = sin(pi/3)`

`=> A-B = pi/3` ..............(ii)

On adding Eqs. (i) and (ii), we get

`2A = (2pi)/3 => A = pi/3` and `B = pi/6`

Correct Answer is `=>` (B) `pi/3`

Q 2450178014

Let `sin (A+ B) = 1` and `sin(A-B) = 1/2` where `AB in [0, pi/2]`

What is the value of `tan(A + 2B) · tan(2A +B)?`
NDA Paper 1 2012

(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

`sin (A+ B) = 1` and `sin(A-B) = 1/2` where `AB in [0, pi/2]`

`=> sin(A+B) = sin(pi/2)`

`=> A+B = pi/2` ...............(i)

and `sin(A-B) = 1/2 => sin(A-B) = sin(pi/3)`

`=> A-B = pi/3` ................(ii)

Now, `tan(A + 2B) · tan(2A + B)`

` = tan(pi/3+pi/3) * tan ((2pi)/3+pi/6)`

` = tan((2pi)/3) * tan((5pi)/6)`

` = tan(pi/2+pi/6) * tan(pi/2+pi/3)`

` = [-cot(pi/6)][-cot(pi/3)] = -(sqrt3) * (-1/sqrt3) = 1`

Correct Answer is `=>` (C) `1`

Q 2470178016

Let `sin (A+ B) = 1` and `sin(A-B) = 1/2` where `AB in [0, pi/2]`

What is the value of `sin^2 A -sin^2 B?`
NDA Paper 1 2012

(A)

`0`

(B)

`1/2`

(C)

`1`

(D)

`2`

Solution:

`sin (A+ B) = 1` and `sin(A-B) = 1/2` where `AB in [0, pi/2]`

`=> sin(A+B) = sin(pi/2)`

`=> A+B = pi/2` ...............(i)

and `sin(A-B) = 1/2 => sin(A-B) = sin(pi/3)`

`=> A-B = pi/3` ..............(ii)

Now, `sin^2 A- sin^2 B= sin^2 (pi/3) - sin^2 (pi/ 6)`

` = (sqrt3/2)^2-(1/2)^2`

`3/4-1/4 = 2/4 = 1/2`

Correct Answer is `=>` (B) `1/2`

Q 2400178018

If `ABCD` is a cyclic quadrilateral, then what is the
value of `sin A + sin B -sin C -sin D?`
NDA Paper 1 2012

(A)

`0`

(B)

`1`

(C)

`2`

(D)

`2(sinA+sinB)`

Solution:

Since, `ABCD` is a cyclic quadrilateral.
Then, `A+ C = 180^0`
and `B + D = 180^0` (by property)
`therefore sin A + sin B -sin C - sin D`
`= sin (180^0- C)+ sin (180^0 - D)- sin C -sinD`
`=sin C + sinD- sin C -sin D`
`=0`

Correct Answer is `=>` (A) `0`

Q 2410278110

What is the value of
`sin 420^0· cos 390^0 + cos (-300^0) · sin(-330^0)?`
NDA Paper 1 2012

(A)

`0`

(B)

`1`

(C)

`2`

(D)

`-1`

Solution:

We have,
`sin 420^0 ·cos 390^0 + cos(- 300^0) ·sin(- 330^0)`

`=sin (360^0 + 60^0) ·cos (360^0 + 30^0) + cos 300^0 (-sin 330^0)`

`= sin 60^0 ·cos 30^0- cos (360^0- 60^0). sin (360^0- 30^0)` `(because cos(-theta) = costheta)`

`=sin 60^0 ·cos 30^0- cos 60^0 · (-sin 30^0)`

`= sin 60^0 · cos 30^0 + cos 60^0 · sin 30^0`

`=sin (60^0 + 30^0) = sin 90^0 = 1`

Correct Answer is `=>` (B) `1`

Q 2450278114

Consider the following statements

I. `1^0` in radian measure is less than `0.02` radians.
II. `1` radian in degree measure is greater than `45^0`.
Which of the above statements is/are correct?
NDA Paper 1 2012

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

I. `1^0 = pi/180` radian

` = 3.14/180 = 0.017 = 0.02 ` (approx)

which is equal to `0.02`

II. 1 Radian ` = (180)/pi` degree = `180/3.14 = 57.32` degree

which is greater than `45^0`.

Correct Answer is `=>` (A) Only I

Q 2480278117

What is the maximum value of `sin^2 x?`
NDA Paper 1 2012

(A)

-1

(B)

`0`

(C)

`1`

(D)

Infinity

Solution:

Let `f(x) = sin^2 x`

`f(x) = 1/2 (2sin^2 x) = 1/2(1-cos2x)`

`f(x) = 1/2-1/2cos2x`

`because -1 le cos2x le 1 => -1/2 le 1/2 cos2x le 1/2`

`=> 1/2 ge -1/2 cos2x ge -1/2`

`=> 1/2+1/2 ge 1/2-1/2cos2x ge 1/2-1/2 => 0le f(x) le1`

Hence, the maximum value of `sin^2 x` is `1`.

Correct Answer is `=>` (C) `1`

Q 2480378217

If `sin theta = cos^2 theta`, then what is `cos^2 theta (1 + cos^2 theta)`
equal to?
NDA Paper 1 2011

(A)

`1`

(B)

`0`

(C)

`cos^2 theta`

(D)

`2 sintheta`

Solution:

Given, `sintheta = cos^2 theta` ... (i)
We have, `cos^2 theta (1 + cos^2 theta)`
`= cos^2 theta + cos^4 theta`
`= 1 - sin^2 theta + cos^4 theta`
`= 1 - (cos^2 theta)^2 + cos^4 theta` [from Eq. (i)]
`= 1- cos^4 theta + cos^4 theta = 1`

Correct Answer is `=>` (A) `1`

Q 2410578419

What is the value of `tan 15^0· tan 195^0`?
NDA Paper 1 2011

(A)

`7-4sqrt3`

(B)

`7+4 sqrt3`

(C)

`7+2sqrt3`

(D)

`7+6sqrt3`

Solution:

`tan 15^0 tan 195^0`

`=tan 15^0 tan (180^0 + 15^0)`

`=tan 15^0·tan 15^0`

`= tan^2 (15^0)`............(i)

`tan2(15^0) = tan30^0 = (2tan15^0)/(1-tan^2 15^0) = 1/sqrt3` `(because tan2theta = (2tan theta)/(1-tan^2 theta))`

`=> 1-tan^2 15^0 = 2 sqrt3 tan15^0`

`=> tan^2 15^0+2sqrt3 tan15^0-1 = 0`

`=> tan15^0 = (-2sqrt3 pm sqrt(12+4)/2) = (-2sqrt3 pm 4)/2`

`=> tan15^0 = (-sqrt3+2)`

From Eq. (i},

`tan^2 15^0 = (2-(sqrt3)^2)`

` = 4+3-4sqrt3`

` = (7-4sqrt3)`

Correct Answer is `=>` (A) `7-4sqrt3`

Q 2460778615

What is `(sinx)/(1+cosx)+(1+cosx)/(sinx)` equal to ?
NDA Paper 1 2011

(A)

`2 tan x`

(B)

`2 cosec x`

(C)

`2 cos x`

(D)

`2 sin x`

Solution:

`(sinx)/(1+cosx)+(1+cosx)/(sinx)`

` = (2sin(x/2)cos(x/2))/(1+2cos^2(x/2)-1)+(1+2cos^2 (x/2)-1)/(2sin(x/2) *cos(x/2))`

` = tan(x/2)+cot(x/2)`

`= (sin(x/2))/(cos(x/2))+(cos(x/2))/(sin(x/2)) = (sin^2(x/2)+cos^2(x/2))/(sin(x/2) * cos(x/2)) * 2/2`

` = 2* 1/sinx = 2cosecx`

Correct Answer is `=>` (B) `2 cosec x`

Q 2420078811

What is the value of
`tan9^0- tan27^0- tan63^0 + tan81^0?`
NDA Paper 1 2011

(A)

`1`

(B)

`2`

(C)

`3`

(D)

`4`

Solution:

`tan 9^0- tan 27^0 - tan 63^0 + tan 81^0`

. `= tan (90^0- 81^0)- tan (90^0- 63^0)- tan 63^0 + tan 81^0`

`=cot 81^0- cot 63^0- tan 63^0 + tan 81^0`

`= (cot 81^0 + tan 81^0)- (cot 63^0 + tan 63^0)`

` = 2/3((cos^2 81^0+sin^2 81^0)/(sin81^0 * cos81^0))-((cos^2 63^0+sin^2 63^0)/(sin63^0 * cos63^0))xx2/2`

` = 2/(sin162^0)-2/(sin126^0) = 2/(sin18^0)-2/(cos36^0)`

` = 2/((sqrt5-1)/4)-2/((sqrt5+1)/4)` `(because sin18^0 = (sqrt5-1)/4 , cos36^0 = (sqrt5+1)/4)`

` = 8{1/(sqrt5-1)-1/(sqrt5+1)} = 8((sqrt5+1-sqrt5+1)/(5-1)) = 8*2/4 = 4`

Correct Answer is `=>` (D) `4`

Q 2480078817

If `x = ycos ((2pi)/3) = z cos ((4pi)/3)` then what is `xy+yz+zx` equal to'?
NDA Paper 1 2011

(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

`x = ycos ((2pi)/3) = z cos ((4pi)/3)`

`=> x = ycos(pi/2+pi/6) = zcos (pi+pi/3)`

`=> x = y{-sin(pi/6)} = z{-cos(pi/3)}`

`=> x = -yxx1/2 = -zxx1/2`

` => 2x = -y = -z`

`=> x/(1/2) = y/(-1) = z/(-1) = k`

`=> x = k/2 , y = -k , z = -k`

Now, `zy + yz + zx = (k/2)(-k)+(-k)(-k)+(-k)(k/2)`

` = (-k^2)/2+k^2-k^2/2 = k^2{(-1+2-1)/2}`

` = (2-2)/2 * k^2 = 0`

Correct Answer is `=>` (B) `0`

Q 2420178911

If `sin A + sin B + sin C = 3`, then what is
`cos A+ cos B +cos C` equal to?
NDA Paper 1 2011

(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`3`

Solution:

Given, `sin A + sin B + sin C = 3`
It is only possible when the `sin A, sin B` and `sin C` has its maximum
values.
i.e., `sin A = sin B = sin C = 1 = sin (pi/2)`

`=> A = B = C = pi/2`

Then, `cos A+ cos B +cos C = cos (pi/2)+cos(pi/2)+cos(pi/2)`

` = 0+0+0 = 0`

Correct Answer is `=>` (B) `0`

Q 2480178917

If `sin3A =1`, then how many distinct values can
`sin A` assume?
NDA Paper 1 2011

(A)

`1`

(B)

`2`

(C)

`3`

(D)

`4`

Solution:

Given, `sin 3A = 1`

`=> 3 sin A - 4 sin^3 A = 1`

`=> 4 sin^3A -3sinA+1 = 0`

`=> sinA = -1 , 1/2 , 1/2`

This is a cubic equation and have three roots in which two roots are same and one different.
So, sin A have 2 distinct values

Correct Answer is `=>` (B) `2`

Q 2440280113

The equation `tan^4 x -2 sec^2 x + a^2 = 0` will have
atleast one real solution, if
NDA Paper 1 2011

(A)

`|a| le 4`

(B)

`|a| le 2`

(C)

`|a| le sqrt3`

(D)

None of these

Solution:

Given equation, `tan^4 x- 2 sec^2 x + a^2 = 0`.

`=> tan^4 x - 2 - 2 tan^2 x + a^2 = 0`

`=> tan^4 x - 2 tan^2 x + (a^2- 2) = 0`

`tan^2x = (2pmsqrt(4-4(a^2-2))/2)`

`=> tan^2 x = (2pm2sqrt(1-a^2+2)/2) = 1pmsqrt(3-a^2)`

For real values of `tan^2 x`,

`3-a^2 le 0 => a^2-3 le 0`

`=> a^2 le 3 => |a| le sqrt3`

Correct Answer is `=>` (C) `|a| le sqrt3`

Q 2440380213

If `tan A- tan B = x` and `cot B -cot A= y`, then
what is `cot (A - B)` equal to?
NDA Paper 1 2011

(A)

`1/y-1/x`

(B)

`1/x-1/y`

(C)

`1/x+1/y`

(D)

`-1/x-1/y`

Solution:

Given, `tan A - tan B = x`
and `cot B - cot A = y`

`=> x = (sinA*cosB-sinB * cosA)/(cosA*cosB) = (sin(A-B))/(cosA cosB)`

`=> 1/x = (cosA * cosB)/(sin(A-B))`

and `y = (cosB * sinA -cosA * sinB)/(sinA*sinB) = (sin(A-B))/(sinA * sinB)`

`=> 1/y = (sinA* sinB)/(sin(A-B))`

`therefore 1/x+1/y = (cosA * cos B+sin A * sinB)/(sin(A-B))`

`=> 1/x+1/y = (cos(A-B))/(sin(A-B))`

`therefore cot(A-B) = 1/x+1/y`

Correct Answer is `=>` (C) `1/x+1/y`

Q 2470480316

If in general, the value of `sin A` is known but the
value of `A` is not known, then how many values of
`tan (A/2)`can be calculated?
NDA Paper 1 2011

(A)

`1`

(B)

`2`

(C)

`3`

(D)

`4`

Solution:

`because sinA = (2tan(A/2))/(1+tan^2 (A/2))`

If `A` is not known but `sin A` is known, then `2` values of `tan (A/2)` can be

calculated because above equation is a quadratic equation in
`tan(A/2)`.

Correct Answer is `=>` (B) `2`

Q 2450680514

If `x = sin B +cos B` and `y = sin B · cos B`, then what
is the value of `x^4 -4x^2 y-2x^2 + 4y^2 + 4y +1?`
NDA Paper 1 2011

(A)

`0`

(B)

`1`

(C)

`2`

(D)

None of these

Solution:

`because x = sintheta+ cos theta` and `y= sin theta * cos theta`

Now, `x^4- 4x^2 y- 2x^2 + 4y^2 + 4y + 1`

`=(sintheta+ cos theta)^4 - 4(sin theta +cos theta )^2 y- 2(sin theta +cos theta)^2 + 4y^2 + 4y + 1`

`= (sin^2 theta + cos^2 theta + 2 sintheta cos theta)^2 - 4(sin^2 theta + cos^2 theta + 2 sintheta cos theta) y-2 (sin^2 theta + cos^2 theta + 2 sin theta cos theta)
+ 4y^2 + 4y + 1`
`= (1 + 2y)^2- 4(1.+ 2y)y-2(1 + 2y)+ 4y^2 + 4y+ 1`
`= 1 + 4y^2 + 4y- 4y - 8y^2- 2- 4y + 4y^2 + 4y + 1 = 0`

Correct Answer is `=>` (A) `0`

Q 2420080811

If `(1 +tan theta)(1 +tanphi) =2,` then what is `(theta + phi)`
equal to?
NDA Paper 1 2011

(A)

`30^0`

(B)

`45^0`

(C)

`60^0`

(D)

`90^0`

Solution:

`(1 + tan theta)(1 + tan phi) = 2`
`=> 1 + tan theta + tan phi + tan theta tan phi = 2`

`=> tan theta + tan phi = 1 - tan theta tan phi`

`=> (tantheta+tanphi)/(1-tantheta tanphi) = 1`

`tan (theta + phi) = tan 45^0`

`(theta+phi) = 45^0`

Correct Answer is `=>` (B) `45^0`

Q 2430180912

If an angle `alpha` is divided into two parts `A` and `B`
such that `A - B = x` and `tan A : tan B = 2 : 1`, then
what is `sin x` equal to?
NDA Paper 1 2011

(A)

`3sin alpha`

(B)

`(2 sin alpha)/3`

(C)

`(sin alpha)/3`

(D)

`2 sin alpha`

Solution:

`because alpha = A+B` and `x = A-B`

`=> A = (alpha+x)/2 , B = (alpha-x)/2`

Now `(tanA)/(tanB) = 2/1 => (sin((alpha+x)/2))/(cos((alpha+x)/2)) * (cos((alpha-x)/2))/(sin((alpha-x)/2)) = 2/1`

` = (2sin((alpha+x)/2) cos((alpha-x)/2))/(2cos((alpha+x)/2) sin((alpha-x)/2)) = 2`

` => (sinalpha+sinx)/(sinalpha-sinx) = 2`

`=> sinalpha+sinx = 2sin alpha-2sinx`

`=> 3sinx = sinalpha`

`=> sinx = (sinalpha)/3`

Correct Answer is `=>` (C) `(sin alpha)/3`

Q 2430291112

What is `(sin theta+1)/(costheta)` equal to ?
NDA Paper 1 2010

(A)

`(sintheta+costheta-1)/(sintheta+costheta+1)`

(B)

`(sintheta+costheta+1)/(sintheta+costheta-1)`

(C)

`(sintheta-costheta-1)/(sintheta+costheta+1)`

(D)

`(sintheta-costheta+1)/(sintheta+costheta-1)`

Solution:

`(1+sintheta)/(costheta) = (1+2tan(theta/2))/(1+tan^2 (theta/2))/(1- tan^2(theta/2))/(1+tan^2(theta/2))`

` = (1+tan(theta/2))^2/((1-tan^2(theta/2)`

` = (1+tan (theta/2))/(1-tan(theta/2)) = (cos(theta/2)+sin(theta/2))/(cos(theta/2)-sin(theta/2))`

Now `(sintheta-costheta+1)/(sintheta+costheta-1) = (2 sin(theta/2) cos (theta/2)+2 sin^2 (theta/2))/(2 sin(theta/2)cos(theta/2)-2sin^2(theta/2))`

` = ( cos(theta/2)+sin(theta/2))/(cos(theta/2)-sin(theta/2))`

Correct Answer is `=>` (D) `(sintheta-costheta+1)/(sintheta+costheta-1)`

Q 2410591410

What is `((sec18^0)/(sec144^0)+(cosec18^0)/(cosec144^0))` equal to ?
NDA Paper 1 2010

(A)

`sec 18^0`

(B)

`cosec 18^0`

(C)

`-sec 18^0`

(D)

`-cosec 18^0`

Solution:

`((sec18^0)/(sec144^0)+(cosec18^0)/(cosec144^0))`

` = (sec18^0)/(sec(180^0-36^0)) +(cosec18^0)/(cosec(180^0-36^0))`

` = -(sec18^0)/(sec36^0)+(cosec18^0)/(cosec36^0) = (sin36^0)/(sin18^0)-(cos36^0)/(cos18^0)`

` = (sin36^0 cos18^0-cos36^0sin18^0)/(sin18^0cos18^0)`

` = (sin(36^0-18^0))/(sin18^0cos18^0) = (sin18^0)/(sin18^0cos18^0) = sec 18^0`

Correct Answer is `=>` (A) `sec 18^0`

Previous Year Questions

Set-2

Q 2460591415

If `alpha` and `beta` are positive angles such that `alpha + beta = pi/4`,
then what is `(1 + tan alpha)(1 + tan beta)` equal to?
NDA Paper 1 2010

(A)

`0`

(B)

`1`

(C)

`2`

(D)

`3`

Solution:

`because alpha+beta = pi/4`

`therefore tan(alpha+beta) = tan(pi/4)`

`=> (tanalpha+tanbeta)/(1-tanalphatanbeta) = 1`

`=> tanalpha+tanbeta = 1-tanalphatanbeta`

`=> 1+tanalpha+tanbeta+tanalphatanbeta = 2`

`=> (1+tanalpha)(1+tanbeta) = 2`

Correct Answer is `=>` (C) `2`

Q 2420691511

What is the value of `(sin 50^0-sin 70^0 +sin 1 0^0)?`
NDA Paper 1 2010

(A)

`1`

(B)

`1/sqrt2`

(C)

`sqrt3/2`

(D)

`0`

Solution:

`(sin 50^0-sin 70^0 +sin 1 0^0)`

` = 2cos((70^0+50^0)/2) * sin((50^0-70^0)/2)+sin10^0 = 0`

` = -2cos60^0sin10^0+sin10^0 = -sin10^0+sin10^0 = 0`

Correct Answer is `=>` (D) `0`

Q 2410791610

If `cos A + cos B = m` and `sin A + sin B = n`, where
`m, n ne 0`, then what is `sin (A+ B)` equal to?
NDA Paper 1 2010

(A)

`(mn)/(m^2+n^2)`

(B)

`(2mn)/(m^2+n^2)`

(C)

`(m^2+n^2)/(2mn)`

(D)

`(mn)/(m+n)`

Solution:

`because cosA+cosB = m` ............(i)

and `sin A+ sin B = n` ... (ii)
From Eqs. (i) and (ii),

`(2mn)/(m^2+n^2) = (2(cosA+cosB)(sinA+sinB))/((cosA+cosB)^2+(sinA+sinB)^2)`

` = (sin2A+sin2B+2sin(A+B))/(1+1+2cos(A-B))`

` = (sin(A+B)[2+2cos(A-B)])/(2+2cos(A-B))`

` = sin(A+B)`

`therefore sin(A+B) = (2mn)/(m^2+n^2)`

Correct Answer is `=>` (B) `(2mn)/(m^2+n^2)`

Q 2430091812

What is the value of `sin 15^0 sin 75^0'?`
NDA Paper 1 2010

(A)

`1/4`

(B)

`1/8`

(C)

`1/16`

(D)

`1`

Solution:

`sin 15^0sin 75^0= sin (45^0- 30^0) sin (45^0 + 30^0)`
`= (sin 45^0 cos 30^0- cos 45^0 sin 30^0)`
`(sin 45^0 cos 30^0 + cos 45^0 sin 30^0)`

` = (1/sqrt2* sqrt3/2 -1/sqrt2 * 1/2)(1/sqrt2* sqrt3/2+1/sqrt2*1/2)`

`((sqrt3-1)/(2sqrt2))((sqrt3+1)/(2sqrt2)) = (3-1)/8 = 2/8 = 1/4`

Correct Answer is `=>` (A) `1/4`

Q 2420191911

What is the value of `(sintheta+costheta-tantheta)/(sectheta+cosectheta-cottheta)` when `theta = (3pi)/4`
NDA Paper 1 2010

(A)

`0`

(B)

`1`

(C)

`-1`

(D)

None of these

Solution:

`(sin((3pi)/4)+cos ((3pi)/4)-tan((3pi)/4))/(sec((3pi)/4)+cosec((3pi)/4)-cot((3pi)/4))`

` = (1/sqrt2-1/sqrt2+1)/(-sqrt2+sqrt2+1) = 1`

Correct Answer is `=>` (B) `1`

Q 2401101028

Which one of the following is correct?
NDA Paper 1 2010

(A)

`sin 1^0 > sin 1`

(B)

`sin 1^0 < sin 1`

(C)

`sin 1^0 =sin 1`

(D)

`sin 1^0 = pi/180 sin1`

Solution:

We know that, `1^0 < 1` rad `=> sin 1^0 < sin 1`

Correct Answer is `=>` (B) `sin 1^0 < sin 1`

Q 2421201121

If `tan A = 1/2` and `tanB = 1/3` then what is the value of `(A+B)?`
NDA Paper 1 2010

(A)

`0`

(B)

`pi/4`

(C)

`pi/2`

(D)

`pi`

Solution:

`because tanA = 1/2` and `tanB = 1/3`

`therefore tan(A+B) = (tanA+tanB)/(1-tanAtanB)`

` = (1/2+1/3)/(1-1/2*1/3) = (5/6)/(5/6) = 1`

` = tan(pi/4)`

`=> A+B = pi/4`

Correct Answer is `=>` (B) `pi/4`

Q 2461201125

If `cos x ne -1`, then what is `(sinx)/(1+cosx)` equal to ?

NDA Paper 1 2010

(A)

`-cot(x/2)`

(B)

`cot(x/2)`

(C)

`tan(x/2)`

(D)

`-tan(x/2)`

Solution:

`(sinx)/(1+cosx) = (2sin(x/2) cos(x/2))/(1+2cos^2(x/2)-1)`

` = (2 sin(x/2) cos(x/2))/(2cos^2(x/2)) = (sin(x/2))/(cos(x/2)) = tan(x/2)`

Correct Answer is `=>` (C) `tan(x/2)`

Q 2401201128

What is the value of `(1+tan15^0)/(1-tan15^0)` ?

NDA Paper 1 2010

(A)

`1`

(B)

`1/sqrt2`

(C)

`1/sqrt3`

(D)

`sqrt3`

Solution:

`(1+tan15^0)/(1-tan15^0) = (tan45^0+tan15^0)/(1-tan45^0tan15^0)` `(tan45^0 = 1)`

`tan(45^0+15^0)`

`tan60^0 = sqrt3`

Correct Answer is `=>` (D) `sqrt3`

Q 2421301221

What is the value of `sqrt3 cosec 20^0- sec 20^0?`
NDA Paper 1 2010

(A)

`1/4`

(B)

`4`

(C)

`2`

(D)

`1`

Solution:

`sqrt3 cosec 20^0- sec 20^0`

` = sqrt3/(sin20^0)-1/(cos20^0) = (sqrt3 cos20^0-sin20^0)/(sin20^0cos20^0)`

`4/(2 sin20^0cos20^0) (sqrt3/2 cos20^0-1/2 sin20^0)`

` = 4/(sin40^0)(sin60^0cos20^0-cos60^0sin20^0)`

` = 4/(sin40^0) sin(60^0-20^0) = 4/(sin40^0) * sin40^0`

` = 4`

Correct Answer is `=>` (B) `4`

Q 2471501426

What is tan `tan(7 1/2)^0` equal to ?
NDA Paper 1 2010

(A)

`sqrt6 + sqrt3 - sqrt2 + 2`

(B)

`sqrt6 + sqrt3 + sqrt2 + 2`

(C)

`sqrt6 - sqrt3 + sqrt2 - 2`

(D)

`sqrt6 + sqrt3 + sqrt2 - 2`

Solution:

`tan(7 1/2)^0 = (2 sin^2 (7 1/2)^0)/(2 sin(7 1/2)^0 cos(7 1/2)^0)`

`( because 2 sin^2(theta/2) = 1-costheta` and `2sin (theta/2) * cos(theta/2) = sintheta)`

` = (1-cos15^0)/(sin15^0)`

` = (1-cos(45^0-30^0))/(sin(45^0-30^0))`

` = (1-(cos45^0*cos30^0+sin45^0sin30^0))/(sin45^0 * cos30^0-cos45^0* sin30^0)`

` = (1- (sqrt3+1)/(2 sqrt2))/((sqrt3-1)/(2 sqrt2))`

` = ((2 sqrt2-sqrt3-1))/(sqrt3-1)*(sqrt3+1)/(sqrt3+1)`

`(2sqrt6-3-sqrt3+2sqrt2-sqrt3-1)/(3-1)`

` = sqrt6-sqrt3+sqrt2-2`

Correct Answer is `=>` (C) `sqrt6 - sqrt3 + sqrt2 - 2`

Q 2411601529

What is the value of `(cos15^0+cos45^0)/(cos^3 15^0+cos^3 45^0) ?`
NDA Paper 1 2010

(A)

`1/4`

(B)

`1/2`

(C)

`1/3`

(D)

None of these

Solution:

`(cos15^0+cos45^0)/(cos^3 15^0+cos^3 45^0)`

` = (cos15^0+cos45^0)/((cos15^0+cos45^0)(cos^2 45^0+cos^2 15^0 - cos45^0 cos15^0))`

`[because a^3+b^3 = (a+b)(a^2-ab+b^2)]`

` = 1/(cos^2 45^0+ cos^2 15^0-cos45^0cos15^0)`

` = 1/((1/sqrt2)^2+cos^2 (45^0-15^0)-1/sqrt2 * cos15^0)`

` = 1/(1/2 +(cos45^0cos30^0+sin45^0sin30^0)^2-(cos15^0)/sqrt2)`

` = 1/(1/2+(sqrt3/(2sqrt2)+1/(2sqrt2))^2-1/sqrt2((sqrt3+1)/(2sqrt2))`

` = 1/(1/2(3+1+2sqrt3)/8-(sqrt3+1)/4)`

` = 1/((4+4+2sqrt3-2sqrt3-2)/8) = 8/6 = 4/3`

Correct Answer is `=>` (D) None of these

Q 2481701627

What is the value of

`| (cos15^0 \ \ \ \ \ \ sin15^0) , (cos45^0 \ \ \ \ sin45^0)| xx | (cos45^0 \ \ \ \ cos15^0) , (sin45^0 \ \ \ \sin15^0)|`

NDA Paper 1 2010

(A)

`1/4`

(B)

`sqrt3/2`

(C)

`-1/4`

(D)

`-3/4`

Solution:

`| (cos15^0 \ \ \ \ \ \ sin15^0) , (cos45^0 \ \ \ \ sin45^0)| xx | (cos45^0 \ \ \ \ cos15^0) , (sin45^0 \ \ \ \sin15^0)|`

` = (sin45^0cos15^0-cos45^0sin15^0) * (cos45^0sin15^0-sin15^0sin45^0cos15^0)`

` = -sin(45^0-15^0) * sin(45^0-15^0)`

` = -sin30^0 * sin30^0`

` = 1/2* 1/2 = -1/4`

Correct Answer is `=>` (C) `-1/4`

Q 2431801722

If the `angle A` lies in the third quadrant and it satisfies
the equation `4 (sin^2 x +cos x) = 1` , then what is the
measure of the `angleA ?`
NDA Paper 1 2010

(A)

`225^0`

(B)

`240^0`

(C)

`210^0`

(D)

None of these

Solution:

`because 4sin^2x+4cosx-1 =0`

`=> 4-4cos^2x+4cos-1 = 0`

`=> -4cos^2x+4cosx+3 = 0`

`=> 4cos^2x-4cosx-3 = 0`

`=> 4cos^2x-6cosx+2cosx-3 = 0`

`=> (2cosx-3)(2cosx+1) = 0`

` => cosx= 3/2`

(which is not possible)

and

`cosx = -1/2`

`=> cosA = -1/2 = cos210^0` (since. A lies in lllrd quadrant)

`therefore A = 210^0`

Correct Answer is `=>` (C) `210^0`

Q 2401001828

What is the value of `tan 15^0 +cot 15^0?`

NDA Paper 1 2009

(A)

`sqrt3`

(B)

`2sqrt3`

(C)

`4`

(D)

`2`

Solution:

`tan15^0+cot15^0 = (sin15^0)/(cos15^0)+(cos15^0)/(sin15^0)`

` = (sin^2 15^0 +cos^2 15^0)/(cos15^0 sin15^0) = 1/(1/2 sin30^0)`

` = 4` ` \ \ \ \ (because sin^2 theta+cos^2 theta = 1)`

Correct Answer is `=>` (C) `4`

Q 2451101924

If `A + B + C = pi/2` then what is the value of

`tan A tan B + tan B tan C + tan C tan A= ?`
NDA Paper 1 2009

(A)

`0`

(B)

`1`

(C)

`-1`

(D)

`tan A tan B tan C`

Solution:

Given `A+B+C = pi/2`

`=> tan(A+B+C) = tan(pi/2) =oo`

` => (tanA+tanB+tanC-tanAtanBtanC)/(1-tanAtanB -tanBtanC-tanCtanA) = 1/0`

`=> tan A tan B + tan B tan C + tan C tan A = 1`

Correct Answer is `=>` (B) `1`

Q 2471112026

If `(sin x +cosec x)^2 +(cos x +sec x)^2
= k + tan^2 x + cot^2 x`, then what is the value of k?
NDA Paper 1 2009

(A)

`8`

(B)

`7`

(C)

`4`

(D)

`3`

Solution:

Given, `(sin x +cosec x)^2 + (cos x +sec x)^2 = k + tan^2 x + cot^2 x`

`=> sin^2 x + cosec^2 x + 2 + cos^2 x + sec^2 x + 2 = k + tan^2 x + cot^2 x`

`=> 1+(cosec^2 x-cot^2 x)+(sec^2 x-tan^2 x)+4 = k` `(because cosec^2 theta = 1+cot^2 theta` and `sec^2 theta = 1+tan^2 theta)`

`=> 1+1+1+4 = k`

`=> k = 7`

Correct Answer is `=>` (B) `7`

Q 2481212127

If `A = (41pi)/12` then what is the value of `(1-3tan^2 A)/(3tanA-tan^3A) ?`

NDA Paper 1 2009

(A)

`-1`

(B)

`1`

(C)

`1/3`

(D)

`3`

Solution:

`(1-3tan^2A)/(3tanA-tan^3A) = 1/(tan3A)`

` = 1/(tan((41pi)/4))` `(because A = (41pi)/12)`

` = 1/(tan(10pi+pi/4)) = 1/tan(pi/4)`

` =1`

Correct Answer is `=>` (B) `1`

Q 2441512423

Consider the following statements
I. If `theta = 1200^0`, then `(sec theta +tan theta )^(-1)` is positive.
of
II. If `theta = 1200^0` , then `{cosec theta -cot theta)` is negative.
Which one of the statements given above is/are correct?
NDA Paper 1 2009

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

I. `(sec theta + tan theta)^(-1) =(sec 1200^0 + tan 1200^0)^(-1)`

`=[sec (6pi + 120^0) + tan (6pi + 120^0)]^(-1)`

`=(sec 120^0 + tan 120^0)^(-1)`

`= (- cosec 30^0 - cot 30^0)^( -1)`

` = (-2-sqrt3)^(-1) = -1/(2+sqrt3)` (negative)

II. `cosec theta-cot theta =cosec (6pi + 120^0)- cot (6pi + 120^0^0)`

`= cosec 120^0 - cot 120^0`

`= sec 30^0 + tan 30^0`

` = 2/sqrt3+sqrt3` (positive)

Hence, both the statements are incorrect

Correct Answer is `=>` (A) Only I

Q 2441612523

If `cottheta = 2 cos theta`, where `(pi/2) < theta < pi` then what is the value of `theta` ?
NDA Paper 1 2009

(A)

`(5pi)/6`

(B)

`(2pi)/3`

(C)

`(3pi)/4`

(D)

`(11pi)/12`

Solution:

Given, `cot theta = 2 cos theta`

`=> cos theta (1 - 2 sin theta) = 0`

For `pi/2 < theta < pi` `costheta ne 0`

`therefore 1-2sintheta = 0 => sintheta = 1/2`

`therefore theta = (5pi)/6`

Correct Answer is `=>` (A) `(5pi)/6`

Q 2461812725

If `cot theta = 5/12`. and `theta` lies in the third quadrant, then

what is `(2 sin theta + 3 costheta)` equal to?
NDA Paper 1 2009

(A)

`-4`

(B)

`-p^2` tor some odd prime `p`

(C)

`(-q/p)` where `p` is an odd prime and `q` is a positive integer with `(q/p)` not an integer

(D)

`-p` for some odd prime `p`

Solution:

Given, `cot theta = 5/12` (since, `theta` iies in Illrd quadrant)

`=> sintheta = -12/13 , costheta = -5/13`

`therefore 2sintheta+3costheta = 2(-12/13)+3(-5/13)`

` = (-24-15)/(13) = (-39)/(13) = -3`

Correct Answer is `=>` (D) `-p` for some odd prime `p`

Q 2441012823

What is the value of

`cos(pi/9)+cos(pi/3)+cos((5pi)/9)+cos((7pi)/9)?`
NDA Paper 1 2009

(A)

`1`

(B)

`-1`

(C)

`-1/2`

(D)

`1/2`

Solution:

`cos(pi/9)+cos(pi/3)+cos((5pi)/9)+cos((7pi)/9)`

` = cos(20^0)+cos(60^0)+{cos(100^0)+cos(140^0)}`

`= cos20^0+1/2-2cos120^0cos20^0`

`= cos20^0+1/2-2sin30^0cos20^0`

` = cos20^0+1/2-cos20^0`

` = 1/2`

Correct Answer is `=>` (D) `1/2`

Q 2401112928

What is the value of `sqrt3 cosec 20^0-sec20^0 ?`
NDA Paper 1 2009

(A)

`4`

(B)

`3`

(C)

`2`

(D)

`1`

Solution:

`sqrt3 cosec20^0-sec20^0 =( sqrt3/sin20^0 ) - (1/cos20^0)`

` = 2((sqrt3/2 cos20^0-1/2 sin20^0)/(sin20^0 * cos20^0))`

` = 2{(sin60^0 * cos20^0-cos60^0* sin20^0)/(sin20^0 * cos20^0)}`

` = 2[sin(60^0-20^0)]/(1/2 sin40^0) = (4 sin40^0)/(sin40^0) = 4`

Correct Answer is `=>` (A) `4`

Q 2431223122

Column I		Column II
(A)	`tan15^0`	(1)	`-2-sqrt3`
(B)	`tan75^0`	(2)	`2+sqrt3`
(C)	`tan105^0`	(3)	`-2+sqrt3`
		(4)	`2-sqrt3`

NDA Paper 1 2009

(A)

`A -> 4 , B -> 1 , C -> 2`

(B)

`A -> 4 , B -> 2 , C -> 1`

(C)

`A -> 3 , B -> 2 , C -> 1`

(D)

`A -> 2 , B -> 1 , C -> 4`

Solution:

`A. tan15^0 = tan(45^0-30^0) = (tan45^0-tan30^0)/(1+tan45^0* tan30^0)`

` = (1-1/sqrt3)/(1+1/sqrt3) = (sqrt3-1)/(sqrt3+1) * (sqrt3-1)/(sqrt3-1)`

` = (3+1-2sqrt3)/2 = 2-sqrt3`

`B. tan75^0 = tan(45^0+30^0) = (tan45^0+tan30^0)/(1-tan45^0*tan30^0)`

` = (1 + 1/sqrt3)/(1-1/sqrt3) = (sqrt3+1)/(sqrt3-1) * (sqrt3+1)/(sqrt3+1) = 2+sqrt3`

`C. tan(105^0) = tan(60^0+45^0) = (tan60^0+tan45^0)/(1-tan60^0*tan45^0)`

` = (sqrt3+1)/(1-sqrt3)*(1+sqrt3)/(1+sqrt3) = (4+2sqrt3)/(-2) = -2-sqrt3`

Correct Answer is `=>` (B) `A -> 4 , B -> 2 , C -> 1`

Q 2471223126

If `P =sin (989^0) cos (991^0)`, then which is the
correct statement in following
NDA Paper 1 2009

(A)

P is finite and positive

(B)

P is finite and negative

(C)

P = 0

(D)

P is not defined

Solution:

Given, `p =sin (989^0). cos (991^0)`

`=sin (1080^0- 91^0) ·cos (1 080^0- 89^0)`

`= - (sin 91^0). (cos 89^0)`

`= -·sin (90^0 + 1) ·cos 89^0=- cos 1^0. cos' 89^0`

Here, `cos 1^0` and `cos 89^0` are positives.
So, Pis rational and negative.
Hence, we can say that P is finite and negative

Correct Answer is `=>` (B) P is finite and negative

Q 2431523422

For which acute angle `theta, cosec^2 theta = 3sqrt3 cot theta - 5 ?`
NDA Paper 1 2009

(A)

`(5pi)/12`

(B)

`pi/3`

(C)

`pi/6`

(D)

`pi/4`

Solution:

`because cosec^2 theta = 3 sqrt3 cot theta -5`

`=> 1+cot^2 theta-3sqrt3 cot theta+5 = 0` `(because cosec^2 theta = 1+cot^2 theta )`

`=> cot^2 theta -3 sqrt3 cot theta+6 = 0`

`cot theta = (3 sqrt3 pm sqrt(27-24))/2 = (3sqrt3 pm sqrt3)/2 = 2 sqrt3 , sqrt3`

`=> cottheta ne 2 sqrt3 , cottheta = sqrt3 = cot(pi/6)`

`therefore theta = pi/6`

Correct Answer is `=>` (C) `pi/6`

Q 2411723620

If `tan^2 theta = 2 tan^2 phi + 1`, then which one of the
following is correct?
NDA Paper 1 2009

(A)

`cos(2theta) = cos(2phi)-1`

(B)

`cos(2theta) = cos(2phi)+1`

(C)

`cos(2theta) = [(cos(2phi)-1)/2]`

(D)

`cos(2theta) = [(cos(2phi)+1)/2]`

Solution:

`cos2phi-1 = (1-tan^2 phi)/(1+tan^2 phi)-1`

` = - (2tan^2 phi)/(1+tan^2 phi)`

` = (-(tan^2 theta -1))/(1+((tan^2 theta-1)/2))` `( because tan^2 theta = 2tan^2 phi+1)`

` = ((1-tan^2 theta)/(1+tan^2 theta)) *2 = cos2theta * 2`

thus `cos 2 theta = (cos2phi-1)/2`

Correct Answer is `=>` (C) `cos(2theta) = [(cos(2phi)-1)/2]`

Q 2411134020

If `cot (x + y) = 1/sqrt3 ` and `cot (x-y) = sqrt3` then what are the smallest positive values of x and y, respectively?
NDA Paper 1 2009

(A)

`45^0, 30^0`

(B)

`30^0 , 45^0`

(C)

`15^0 , 60^0`

(D)

`45^0 , 15^0`

Solution:

`because cot(x+y) = 1/sqrt3 = cot60^0`

`=> x+y = 60^0` ...........(i)

and `cot (x-y) = sqrt3 = cot30^0`

`=> x-y = 30^0` ..........(ii)

From Eqs. (i) and (ii),

`x = 45^0` , and `y = 15^0`

Correct Answer is `=>` (D) `45^0 , 15^0`

Q 2461134025

If `sin A =1/sqrt5` and `cos B = 3/sqrt(10)`,where `A` and `B`

being positive acute angles, then what is `(A+ B)`
equal to?
NDA Paper 1 2009

(A)

`pi/6`

(B)

`pi/4`

(C)

`pi/3`

(D)

`pi/2`

Solution:

`sin A =1/sqrt5` and `cos B = 3/sqrt(10)`

`therefore sin(A+B) = sinAcosB+cosAsinB`

` = 1/sqrt5 * 3/sqrt(10)+sqrt(1-1/5) *sqrt(1-9/10)`

` = 3/sqrt(50)+2/sqrt5*1/sqrt(10) = (3+2)/sqrt(50) = 1/sqrt2 = sin(pi/4)`

`=> A+B = pi/4`

Correct Answer is `=>` (B) `pi/4`

Q 2431634522

What is the value of
`cos 10^0 +cos 110^0 +cos 130^0?`
NDA Paper 1 2009

(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

`cos 10^0 + cos 110^0 + cos 130^0`

`=(cos 130^0 +cos 10^0) +cos 110^0`

`= 2 cos 60^0 cos 70^0 + cos 11 0^0`

`=cos 70^0 + cos 110^0= cos (180^0- 110^0) + cos 110^0`

`= - cos 11 0^0 + cos 11 0^0 = 0`

Correct Answer is `=>` (B) `0`

Q 2411034820

What is `sqrt(2+sqrt2+sqrt2+sqrt(2+2cos4A))` equal to ?
NDA Paper 1 2008

(A)

`cosA`

(B)

`cos(2A)`

(C)

`2cos(A/2)`

(D)

`sqrt(2cosA)`

Solution:

`sqrt(2+sqrt2+sqrt2+sqrt(2+2cos4A))`

` = sqrt(2+sqrt2+sqrt2+sqrt(2(1+cos4A)))`

`(because cos4A+1 = 2cos^2 2A)`

`= sqrt(2+sqrt2+sqrt(4cos^2 2A))`

` = sqrt(2+sqrt(2+2cos2A))`

` = sqrt(2+sqrt(2(1+cos2A))` `(because cos2A +1 = 2cos^2 A)`

` = sqrt(2+sqrt(4cos^2 A))`

` = sqrt(2+2cosA))`

`= sqrt(2(1+cosA)) ` `(because A+1 = 2cos^2(A/2))`

`= sqrt(4 cos^2 (A/2)) = 2 cos (A/2)`

Correct Answer is `=>` (C) `2cos(A/2)`

Q 2401034828

What is the value of `sin(1110^0)?`
NDA Paper 1 2008

(A)

`1`

(B)

`1/2`

(C)

`1/sqrt2`

(D)

`sqrt3/2`

Solution:

`Sin (111 0^0) = Sin { 3 xx 360^0 + 30^0} = Sin 30^0 = 1/2`

Correct Answer is `=>` (B) `1/2`

Q 2461145025

The equation `tan^2 phi+tan^6 phi = tan^3 phi * sec^2 phi` is
NDA Paper 1 2008

(A)

identity for only one value of `phi`

(B)

not an identity

(C)

identity for all values of `phi`

(D)

None of the above

Solution:

Given that, `tan^2 phi + tan^6 phi = tan^3 phi * sec^2 phi`

`=> tan^2 phi{1 + tan^4 phi- tan phi . sec^2 phi} = 0`

`=> tan^2 phi {1 + tan^4 phi - tan phi - tan^3 phi} = 0`

`=> tan^2 phi{(1 - tanphi)+ tan^3 phi (tan phi - 1)} = 0`

`=> tan^2 phi{(1 - tan phi)(1 - tan^3 phi)} = 0`

`=> tan^2 phi (1 - tan phi)^2 (1 + tan^2 phi+ tan phi) = 0`

Hence, it is notation identity for all value of `phi` because at
`phi= (2n + 1) pi/2` the given identify does not exist.

Correct Answer is `=>` (D) None of the above

Q 2411345220

What is the value of `cos 15^0?`
NDA Paper 1 2008

(A)

`1/2(sqrt(2-sqrt3))`

(B)

`1/2(sqrt(2+sqrt3))`

(C)

`sqrt2+sqrt3`

(D)

`sqrt2-sqrt3`

Solution:

`cos2theta = 2cos^2 theta-1`

put `theta = 15^0`

`therefore cos30^0 = 2cos^2 15^0-1`

`=> sqrt3/2+1 = 2cos^2 15^0`

`=> cos^2 15^0 = (sqrt3+2)/4`

`therefore cos15^0 = 1/2 sqrt(2+sqrt3)`

Alternate Method
`cos 15^0 =cos (45^0- 30^0)`

`=cos 45^0·cos 30^0 + sin 45^0 ·sin 30^0`

` = 1/sqrt2 * sqrt3/2+1/sqrt2 *1/2 = (sqrt3+1)/(2sqrt2) = (sqrt6+sqrt2)/4`

` = [(sqrt(sqrt6+sqrt2))]^2 = sqrt(4+4sqrt3)/4`

` = 2sqrt(2+sqrt3)/4 = 1/2 sqrt(2+sqrt3)`

Correct Answer is `=>` (B) `1/2(sqrt(2+sqrt3))`

Q 2431345222

How many values of `theta` between `0^0` and `360^0` satisfy
`tan theta = k ne 0`, where `k` is a given number?
NDA Paper 1 2008

(A)

`1`

(B)

`2`

(C)

`4`

(D)

Many

Solution:

When `k` is a given number, then the value of `theta` between
`0^0` and `360^0` satisfy `tan theta = k ne 0` is many because only at `theta = pi`
`=> tan theta = 0^0`

Correct Answer is `=>` (D) Many

Q 2461345225

NDA Paper 1 2008

Assertion : Let `X= {theta = [0, 2pi] sintheta = cos theta}` (A) The number of elements in `X` is `2`.

Reason : (R) `sin theta` and `cos theta` are both negative both in second are fourth quadrants

(A)

Both A and R individually true and R is the correct explanation of A

(B)

Both A and R are individually true but R is not the correct explanation of A

(C)

A is true but R is false

(D)

A is false but R is true

Solution:

`because X = { theta = [ 0 , 2pi] : sintheta = costheta}`

Thus, the number of elements in `X` is `2`. Since, `sin theta = cos theta` is
possible at `theta = 45^0` and `225^0`.
While `sin theta` positive, negative in Ilind, IVth quadrant respectively
and `cos theta` positive, negative in IVth, lind quadrant respectively.
Hence, A is true but R is false.

Correct Answer is `=>` (C)

Q 2411445320

What is the measure of the angle `114^0 35' 30''` in
radian?
NDA Paper 1 2008

(A)

1 rad

(B)

2 rad

(C)

3 rad

(D)

4 rad

Solution:

We know that, `60" = 1' => 30" = 1"/2`

`therefore 35' 30 = (30+1/2)' = (71/2)` and `60' = 1^0`

`therefore 1' = (1/60)^0`

`=> (71/2)' = (71/2xx1/60)^0 = (71/120)^0`

`therefore 114^0 35' 30" = (114+71/120)^0 = (13751/120)^0`

We know that, `2pi` rad `= 360^0`

`=> (13751/120)^0 = (2pi)/(360^0)xx(13751)/(120)` rad

` = (2xx22xx13751)/(7xx360xx120)` rad

` = 2.0008069` rad

`therefore 114^0 35' 30" = 2` rad (approx)

Correct Answer is `=>` (B) 2 rad

Q 2451645524

Which one of the following is correct?

`(1+cos67 1^0/2)(1+cos112 1^0/2)` is

NDA Paper 1 2008

(A)

an irrational number and is greater than 1

(B)

a rational number but not an integer

(C)

an integer

(D)

an irrational number and is less than 1

Solution:

`(1+cos67 1^0/2)(1+cos112 1^0/2)`

`= (1+cos67 1^0/2){1+cos(180^0-112 1^0/2)}`

` = (1+cos 67 1^0/2)(1-cos 67 1^0/2)`

` = 1-cos^2 67 1^0/2 = sin^2 67 1^0/2`

` = (1-cos135^0)/2 = (sqrt2+1)/(2sqrt2)` `(becausecos135^0 = -1/sqrt2)`

which is an irrational number and is less than 1 .

Correct Answer is `=>` (D) an irrational number and is less than 1

Q 2411645529

If `sin2A == 4/5` then what is the value of `tanA \ \ \ (` where, ` 0 le A le pi/4)?`

NDA Paper 1 2008

(A)

`1`

(B)

`-1`

(C)

`1/2`

(D)

`2`

Solution:

`because sin2A = 4/5`

`therefore (2tanA)/(1+tan^2A) = 4/5`

`=> 10 tanA = 4+4tan^2A`

`=> 4tan^2A-8tanA-2tan+4 =0`

`=> 2tanA(2tanA-4)-2tanA-2) = 0`

`=> (2tanA-1)(tan-2) = 0`

`tanA = 1/2` `(because tanA ne 2)`

Correct Answer is `=>` (C) `1/2`

Q 2421745621

What is the value of `(cos10^0-sin10^0)/(cos10^0+sin10^0)?`
NDA Paper 1 2008

(A)

`tan35^0`

(B)

`tan10^0`

(C)

`1/sqrt2`

(D)

`1`

Solution:

`(cos10^0-sin10^0)/(cos10^0+sin10^0)`

` = (1-tan10^0)/(1+tan10^0)`

` = (tan45^0-tan10^0)/(1+tan45^0tan10^0)`

` = tan(45^0-10^0)`

`= tan35^0`

Correct Answer is `=>` (A) `tan35^0`

Q 2441745623

For what value of `x` does the equation
`4 sin x + 3 sin 2x -2 sin 3x +sin 4x =2sqrt3` hold?
NDA Paper 1 2008

(A)

`pi/6`

(B)

`pi/4`

(C)

`pi/3`

(D)

`pi/2`

Solution:

`because 4 sin x + 3 sin 2x -2 sin 3x +sin 4x =2sqrt3 `

Let `x = pi/6`

`therefore 4sin(pi/6)+3sin(pi/3) - 2sin(pi/2)+sin((2pi)/3)`

` = 4 1/2+(3sqrt3)/2-2+sqrt3/2`

` = 2sqrt3`

`therefore x = pi/6`

Correct Answer is `=>` (A) `pi/6`

Q 2411745629

Which one of the following pairs is not correctly
matched?
NDA Paper 1 2008

(A)

`sin2pi : sin(-pi)`

(B)

`tan45^0 : tan(-315^0)`

(C)

`cot(tan^(-1) 0.5) : tan(cot^(-1) 0.5)`

(D)

`tan420^0 : tan(-60^0)`

Solution:

`sin2pi ne sin(-pi)`

`because sin(-2pi) = -sin2pi`

Correct Answer is `=>` (A) `sin2pi : sin(-pi)`

Q 2411145929

What is the value of `sin((5pi)/12)`?
NDA Paper 1 2008

(A)

`(sqrt3+1)/2`

(B)

`(sqrt6+sqrt2)/4`

(C)

`(sqrt3+sqrt2)/4`

(D)

`(sqrt6+1)/2`

Solution:

`sin \ (5pi)/(12) = sin75^0`

` = sin(45^0+30^0)`

` = sin45^0 cos30^0+cos45^0sin35^0`

` = 1/sqrt2*sqrt3/2+1/sqrt2*1/2`

` = (sqrt3+1)/(2sqrt2) * sqrt2/sqrt2`

` = (sqrt6+sqrt2)/4`

Correct Answer is `=>` (B) `(sqrt6+sqrt2)/4`

Q 2411156020

What is the correct sequence of the following
values?

`I. sin(pi/12)`

`II. cos(pi/12)`

`III. cot(pi/12)`
NDA Paper 1 2008

(A)

Ill > II > I

(B)

I > II > Ill

(C)

I > Ill > II

(D)

Ill > I > II

Solution:

`I. sin(pi/12) = sin15^0 = (sqrt3-1)/(2 sqrt2)`

` because sin15^0 = sin(45^0-30^0)`

` = sin45^0 * cos30^0 -cos45^0 * sin30^0`

` = sqrt3/(2sqrt2)-1/(2sqrt2) = (sqrt3-1)/(2sqrt2)`

`II. cos(pi/12) = cos15^0 = 2+sqrt3`

`because cos15^0 = cos(45^0-30^0)`

` = cos45^0 * sin30^0-cos45^0* sin30^0`

` = sqrt3/(2sqrt2)+1/(2sqrt2) = (sqrt3+1)/(2 sqrt2)`

`III. cot(pi/12) = cot15^0 = 2+sqrt3`

`because cot15^0 = tan75^0 = tan(45^0+30^0)`

` = (tan45^0+tan30^0)/(1-tan45^0*tan30^0)`

` = (1+1/sqrt3)/(1-1/sqrt3)`

` = (sqrt3+1)/(sqrt3-1)`

` = 2+sqrt3`

So, the correct sequence is Ill > II > I.

Correct Answer is `=>` (A) Ill > II > I

Q 2421156021

If `alpha` and `beta` are such that `tan alpha = 2 tan beta`, then what
is `sin (alpha + beta)` equal to?
NDA Paper 1 2007

(A)

`1`

(B)

`2sin(alpha-beta)`

(C)

`sin(alpha-beta)`

(D)

`3sin(alpha-beta)`

Solution:

Given that, `tan alpha = 2 tan beta`

`(sinalpha/cosalpha)/(sinbeta/cosbeta) = 2`

`=> (sinalphacosbeta)/(cosalphasinbeta) = 2/1`

On using componendo and dividendo rule, we get

`(sinalphacosbeta+cosalphasinbeta)/(sinalphacosbeta-cosalphasinbeta) = (2+1)/(2-1)`

` => (sin(alpha+beta))/(sin(alpha-beta)) = 3`

`=> sin(alpha+beta) = 3(sin(alpha-beta))`

Correct Answer is `=>` (D) `3sin(alpha-beta)`

Q 2431156022

What is the value of

`cos 306^0 +cos 234^0 +cos 162^0 +cos 18^0?`
NDA Paper 1 2007

(A)

`1`

(B)

`-1`

(C)

`0`

(D)

`2`

Solution:

`cos 306^0 +cos 234^0 +cos 162^0 +cos 18^0`

`=cos (360^0- 54^0) +cos (180^0 + 54^0)+cos (180^0- 18^0) +cos 18^0`
`=cos 54^0- cos 54^0- cos 18^0 +cos 18^0= 0`

Correct Answer is `=>` (C) `0`

Q 2401156028

If `0^0 < 9 < 45^0`, then which one of the following is
correct?
NDA Paper 1 2007

(A)

`sin^2 theta + cos^6 theta = sin^6 theta + cos^2 theta`

(B)

`cosec^2 theta + cot^6 theta = cosec^6 theta + cot^2 theta`

(C)

`sin^2 theta - cos^4 theta = sin^4 theta + cos^2 theta`

(D)

`cosec^2 theta + cot^4 theta = cosec^4 theta + cot^2 theta`

Solution:

(a) `sin^2 theta + cos^6 theta = sin^6 theta + cos^2 theta`
`=> sin^6 theta - cos^6 theta = sin^2 theta - cos^2 theta`
`sin^6 theta - cos^6 theta = (sin^2 theta - cos^2 theta)`
`(sin^4 theta + cos^4 theta - sin^2 theta cos^2 theta)`
`= (sin^2 theta - cos^2 theta)(1 - 3 sin^2 theta cos^2 theta)`
which is not equal to `sin^2 theta - cos^2 theta`.
(b) `because cosec^6 theta- cot^6 theta`
`= (cosec^2 theta- cot^2 theta)[(cosec^2 theta - cot^2 theta)^2 +cosec theta cottheta]`
which is not equal to `cosec^2 theta- cot^2 theta`.
(c) `sin^4 theta + cos^4 theta = (sin^2 theta + cos^2 theta)^2 - 2 sin^2 theta cos^2 theta`
`= 1 - 2 sin^2 theta cos^2 theta`
which is not equal to `sin^2 theta - cos^2 theta`.
(d) `cosec^2 theta + cot^4 theta = cosec^2 theta + (cosec^2 theta- 1)^2`
`= cosec^2 theta + cosec^4 theta + 1-2 cosec^2 theta`
`= cosec^4 theta + 1- cosec^2 theta =cosec^4 theta + cot^2 theta`

Correct Answer is `=>` (D) `cosec^2 theta + cot^4 theta = cosec^4 theta + cot^2 theta`

Q 2411256120

lf `sin A =sin B` and `cos A =cos B`, then which one
of the following is correct?
NDA Paper 1 2007

(A)

`B = npi + A`

(B)

`A = 2npi - B`

(C)

`A = 2npi + B`

(D)

`B = npi - A`

Solution:

Given that, `sin A = sin B` and `cos A =cos B`

`therefore sin(A-sinB = 0` and `cosA-cosB = 0`

` => 2 sin((A-B)/2) * cos((A+B)/2) = 0`

` => 2 sin((A-B)/2) * sin((A+B)/2) = 0`

`=> 2 sin((A-B)/2) * cos((A+B)/2) = - 2 sin((A-B)/2) * sin((A+B)/2)`

`=> 2sin ((A-B)/2) { sin((A+B)/2)+cos((A+B)/2)} = 0 , sin((A-B)/2) = sin0^0`

` => (A-B)/2 = npi` `( because sin((A+B)/2)+cos((A+B)/2) ne 0)`

`therefore A = 2npi+B`

Correct Answer is `=>` (C) `A = 2npi + B`

Q 2421256121

If `alpha = pi/8` then what is the value of

`cos alpha cos 2alpha cos 4alpha,?`
NDA Paper 1 2007

(A)

`0`

(B)

`1/4`

(C)

`8`

(D)

`4`

Solution:

`alpha = pi/8`

`cos alpha cos 2alpha cos 4alpha = cos(pi/8)*cos(pi/4) * cos(pi/2)` `(because cos(pi/2) = 0)`

` = 0`

Correct Answer is `=>` (A) `0`

Q 2451256124

If `A, B` and `C` are angles of a triangle such that
`tan A = 1, tan B = 2`, then what is the value of
`tan C= ?`
NDA Paper 1 2007

(A)

`0`

(B)

`1`

(C)

`2`

(D)

`3`

Solution:

`because tanA = 1 , tanB = 2` and `A+B+C = 180^0`

We know that

`tan(A+B+C) = (tanA+tanB+tanC - tanAtanBtanC)/(1-tanA tanB - tanBtanC-tanCtanA)`

`=> tan180^0 = (1+2+tanC-2tanC)/(1-2-2tanC-tanC) = 0` `(because 180^0 = 0)`

`=> 3-tanC = 0 => tanC = 3`

Correct Answer is `=>` (D) `3`

Q 2461256125

What is the value of

`(cosec(pi+theta) cot{((9pi)/2)-theta} cosec^2 (2pi-theta))/(cot(2pi-theta) sec^2(pi-theta) sec{((3pi)/2)+theta})`

NDA Paper 1 2007

(A)

`0`

(B)

`1`

(C)

`-1`

(D)

`oo`

Solution:

`(cosec(pi+theta) cot{((9pi)/2)-theta} cosec^2 (2pi-theta))/(cot(2pi-theta) sec^2(pi-theta) sec{((3pi)/2)+theta})`

`[ because cot((9pi)/2 - theta) = cot (pi/2-theta ) = tantheta ]`

`= ((-cosec theta)(tantheta)(cosec^2 theta))/((-cot theta ) (sec^2 theta )(cosec theta))`

` = (tan^2 theta * cosec^2 theta)/(sec^2 theta) = (sin^2 theta)/(cos^2 theta) * (cos^2 theta)/(sin^2 theta) = 1`

Correct Answer is `=>` (B) `1`

Q 2471256126

What is the value of
`sin (A + B) sin (A -B) + sin (B + C) sih (B -C)+sin (C + A) sin (C- A)?`
NDA Paper 1 2007

(A)

`0`

(B)

`sin A + sin B + sin C`

(C)

`cos A + cos B + cos C`

(D)

`1`

Solution:

`because sin (A + B) sin (A - B)`

` = 1/2 { 2 sin(A+B) * sin(A-B)`

` = 1/2 { cos (A-B-A- B)- cos (A- B + A+ B)}`

` = 1/2{cos 2B- cos 2A}`

Similarly, `sin (B +C) sin (B- C = 1/2 {cos 2C -cos 2B}`

and `sin (C + A)· sin (C- A) = 1/2 {cos 2A- cos 2C}`

`therefore sin (A+ B) sin (A- B)+ sin (B + C) sin (B C) + sin(C A)sin(C -·A)`

` = 1/2 cos 2C -cos 2B +cos 2A- cos 2C +cos 2B- cos 2A)`

` = 0`

Correct Answer is `=>` (A) `0`

Q 2401256128

If `tan alpha = m/(m+1)` and `tanbeta = 1/(2m+1)` then what is the value of `alpha + beta?`
NDA Paper 1 2007

(A)

`0`

(B)

`pi/4`

(C)

`pi/6`

(D)

`pi/3`

Solution:

`tan alpha = m/(m+1)` and `tanbeta = 1/(2m+1)` (given)

Now `tan(alpha+beta) = (tan alpha+tanbeta)/(1-tan alpha tanbeta) = (m/(m+1) +1/(2m+1) )/(1- m/(m+1) * 1/(2m+1))`

` = (2m^2+2m+1)/(2m^2+2m+1) = 1`

` = tan(pi/4)`

`therefore alpha+beta = pi/4`

Correct Answer is `=>` (B) `pi/4`

Q 2421456321

What is the value of `cosec((3pi)/12)?`
NDA Paper 1 2007

(A)

`sqrt6+sqrt2`

(B)

`-sqrt6+sqrt2`

(C)

`sqrt6-sqrt2`

(D)

`-sqrt6-sqrt2`

Solution:

`cosec((3pi)/12) = cosec195^0`

` = cosec(180^0+15^0)`

`= -cosec15^0`

` = -sqrt(1+cot^2 15^0)`

`[because cot15^0 = cot(45^0-30^0)]`

`cot(45^0-30^0)= (cot45^0* cot30^0+1)/(cot30^0-cot45^0)`

` = (sqrt3+1)/(-1+sqrt3) * (1+sqrt3)/(1+sqrt3)`

` = (sqrt3+1)^2/(3-1)`

` = (2(2+sqrt3))/2`

` = 2+sqrt3`

` = - sqrt(1+(2+(sqrt3)^2))`

` = -sqrt(1+4+3+4 sqrt3)`

` = - sqrt(8+4 sqrt3)`

` = -sqrt( (sqrt6)^2+(sqrt2)^2+2(sqrt6)(sqrt2))`

` = - sqrt((sqrt6+sqrt2)^2)`

` = -sqrt6-sqrt2`

Correct Answer is `=>` (D) `-sqrt6-sqrt2`

Q 2411656520

If `alpha+beta = pi/2` and `beta+gamma = alpha` then which one of the following is correct?
NDA Paper 1 2007

(A)

`2 tan beta+tangamma = tanalpha`

(B)

`tanbeta+2tangamma = tanalpha`

(C)

`tanbeta+tangamma = tanalpha`

(D)

`2(tanbeta+tangamma) = tanalpha`

Solution:

`because alpha+beta = pi/2` and `beta+gamma = alpha`

`tan(beta+gamma) = tan alpha`

`=> (tanbeta+tangamma)/(1-tanbetatangamma) = tanalpha`

`=> tan beta+tangamma = tanalpha-tanbeta tangamma`

` = tanalpha-tan alpha * tan (pi/2-alpha) * tangamma`

`=> tanbeta+tangamma = tanalpha-tanalphacotalphatangamma` `(because beta+alpha = pi/2)`

` => tanbeta+tangamma = tanalpha-tangamma`

`=> tanbeta+2tangamma = tanalpha`

Correct Answer is `=>` (B) `tanbeta+2tangamma = tanalpha`