Mathematics Tricks & Tips of Basic Trigonometry FOR NDA PART

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Tricks & Tips of Basic Trigonometry FOR NDA PART - 1

Relation between Degrees, Grades and Radians

The relation between the three systems of measurement of an angle is `("Degree")/90 =("Grade")/100 = (2 " Radian")/(pi)` Thus,

(i) To convert radians into degrees multiply by `(180/pi)`

(ii) To convert degrees into radians multiply by `(pi/180)`

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 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`

Trigonometric Ratios of Allied Angles and Fundamental Trigonometric Identities

The following are some fundamental identities

(i) `sin^2 theta+ cos^2 theta=1`
(ii) `1+ tan^2 theta= sec^2 theta`
(iii) `1+ cot^2 theta= cosec^2 theta`

Q 2460523415

What is `sin^2 20^0 + sin^2 70^0` equal to?
NDA Paper 1 2013

(A)

`1`

(B)

`0`

(C)

`-1`

(D)

`1/2`

Solution:

given `sin^2 20^0 + sin^2 70^0`

` = sin^2 20^0+sin^2 (90^0-20^0)`

` = sin^2 20^0+cos^2 20^0 = 1` `(because sin^2 theta+cos^2 theta = 1)`

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

Q 1628178901

`(1 -sin A +cos A)^2` is equal to
NDA Paper 1 2015

(A)

`2 (1 -cos A)(1 +sin A)`

(B)

`2 (1- sin A)(1 +cos A)`

(C)

`2 (1- cos A)(l- sin A)`

(D)

None of the above

Solution:

`(1- sin A+ cosA)^2`

`= 1 + sin^2 A+ cos^2 A -2sinA -2sinA cosA + 2cosA`

`= 2 - 2sinA- sin2A + 2cosA`

`= 2(1 + cos A)- 2sinA(1 +cos A)`

`= 2(1- sinA)(1 + cos A)`

Correct Answer is `=>` (B) `2 (1- sin A)(1 +cos A)`

Q 1688178907

What is ` (cos theta)/( 1 - tan theta) + (sin theta)/( 1 - cot theta)` equal to ?

NDA Paper 1 2015

(A)

`sin theta - cos theta`

(B)

`sin theta + cos theta`

(C)

` 2 sin theta`

(D)

`2 cos theta`

Solution:

` (cos theta)/( 1 - tan theta) + (sin theta)/( 1 - cot theta)`

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

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

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

Correct Answer is `=>` (B) `sin theta + cos theta`

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 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 1712445339

If `A+ B + C = pi`, then what is `cos (A+ B) + cos C` equal
to?
NDA Paper 1 2014

(A)

`0`

(B)

`2 cos C`

(C)

`cos C - sin C`

(D)

`2 sin C`

Solution:

Given that, `A+ B + C = pi ....... (i)`

Now, we have `cos (A+ B) + cos C`

` = cos (pi - C) + cos C` [from Eq. (i)]

`= - cos C + cos C`

`= 0 ( ∵ ` in second quadrant, `cos theta < 0)`

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

Q 1710801710

What is ` sqrt(1 + Sin 2 theta )` equal to?
NDA Paper 1 2014

(A)

`cos theta - sin theta`

(B)

`cos theta + sin theta`

(C)

`2 cos theta + sin theta`

(D)

`cos theta + 2sin theta`

Solution:

Consider, `sqrt(1 + Sin 2 theta )`

` = sqrt( Sin ^2 theta + cos ^2 theta + 2 sin theta cos theta)`

` = sqrt(( Sin theta + cos theta))^2 = Sin theta + cos theta`

Correct Answer is `=>` (B) `cos theta + sin theta`

Q 2411223129

What is the value of `tan (-1575^0)?`
NDA Paper 1 2009

(A)

`1`

(B)

`1/2`

(C)

`0`

(D)

`-1`

Solution:

`tan (-1575^0) = - tan ( 4 xx 360^0 + 135^0) = - tan135^0`

`=-tan (90^0 + 45^0)= cot 45^0 = 1`

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

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 2480834717

What is the value of `sec^2 tan^(-1) (5/11)` ?
NDA Paper 1 2012

(A)

`121/96`

(B)

`217/921`

(C)

`146/121`

(D)

`267/121`

Solution:

Given, `sec^2 tan^(-1)(5/11) ` `(∵ 1+tan^2 theta=sec^2 theta)`

`1+tan^2tan^(-1)(5/11)=1+{tan tan^(-1) (5/11)}^2`

`=1+(5/11)^2=1+25/121=146/121`

Correct Answer is `=>` (C) `146/121`

Q 1742145033

What is value of `( 1 + sin A)/ ( 1 - sin A) - ( 1 - sin A)/( 1 + sin A) ?`
NDA Paper 1 2014

(A)

`sec A- tan A`

(B)

`2 sec A . tan A`

(C)

`4 sec A . tan A`

(D)

`4 cosec A. cot A`

Solution:

`( 1 + sin A)/ ( 1 - sin A) - ( 1 - sin A)/( 1 + sin A)`

` = ( (1 + sin A)^2 - (1 -sin A)^2)/((1 -sin A) (1 + sin A))`

` = (( 1 + sin^2A +2 sin A) - (1 + sin^2 A -2 sin A))/(1 -sin^2 A)`

` (∵(a+ b)^2 = a^2 + 2ab + b^2),(a - b)^2 = a^2 - 2ab + b^2),`

`( (a- b) (a + b) = (a^2 - b^2)) `

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

` (∵ sin^2 A+ cos^2 A = 1)`

` = ( 4 sin A)/( cos^2 A) = 4 (sin A)/(cos A) . 1/(cos A)`

` = 4 tan A . sec A`

` (∵ tan A = (sin A)/(cos A) ` and ` sec A = 1/(cos A)) `

Correct Answer is `=>` (C) `4 sec A . tan A`

Q 2431256122

What is the value of `cot (-870^0)?`
NDA Paper 1 2007

(A)

`sqrt3`

(B)

`1/sqrt3`

(C)

`-sqrt3`

(D)

`-1/sqrt3`

Solution:

`cot (- 870^0) = -cot (2 xx 360^0 + 150^0)`

`= -cot 150^0 = -cot (90^0 + 60^0)`

`=tan 60^0 = sqrt3`

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

Q 2231612522

If `p =tan(- (11pi)/6) , q = tan ((21pi)/4 )` and `r =cot ((283 pi)/6 )`
then which of the following is/are correct?
`1.` The value of `p xx r` is `2`.

2. `p, q` and `r` are in `GP`.

Select the correct answer using the code given below.
NDA Paper 1 2015

(A)

Only `1`

(B)

Only `2`

(C)

Both `1` and `2`

(D)

Neither `1` nor `2`

Solution:

`p = tan (- (11pi)/6) = - tan ((11pi)/6)`

` = - tan (2pi - pi/6) = tan( pi/6) = 1/sqrt(3)`

` = q = tan ((21 pi)/4) = tan ( 5 pi + pi/4) = tan( pi/4) = 1`

` r = cot ((283pi)/6) = cot ( 47 pi +pi/6) = cot(pi/6 )= sqrt(3)`

1. `p xx r = 1/sqrt(3) xx sqrt (3) = 1` which is incorrect.

2. `p, q` and `r` are in `GP`.

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

Q 1732745632

What is `sin^2 (3 pi) + cos^2 (4 pi) + tan^2 (5 pi)` equal to?
NDA Paper 1 2014

(A)

`0`

(B)

`1`

(C)

`2`

(D)

`3`

Solution:

We have,

`sin^2 (3pi) + cos^2 (4pi) + tan^2 (5pi)`

`= sin^2 (2pi + pi) + cos^2 (2pi + 2pi) + tan^2 (4pi + pi)`

` = sin^2 pi + cos^2 2pi + tan^2 pi`

`[∵ sin (2pi + theta) = sin theta , cos(2pi + theta) = cos theta`,

` tan (4pi + theta )= tan theta]`

` = sin^2 ( pi/2 + pi/2 ) + cos^2 ( (3pi)/2 + pi/2 ) + tan^2 ( pi/2 + pi/2 )`

`( ∵sin (pi/2 + theta ) = cos theta ,cos ( (3pi)/2 + theta ) = sin theta , tan (pi/2 + theta ) = - cot theta )`

` = cos^2 (pi/2) + sin^2 (pi/2) + cot^2 (pi/2)`

` = (0)^2 + (1)^2 + (0)^2`

` = 0 + 1 + 0 = 1`

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

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 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 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 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 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 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 2441001823

If `sin x + cosec x = 2`, then what· is the value of
`sin^4 x + cosec^4 x ?`
NDA Paper 1 2009

(A)

`2`

(B)

`4`

(C)

`8`

(D)

`16`

Solution:

Given, `sin x +cosec x = 2`

`therefore sin^4x+cosec^4x = (sin^2x+cosec^2 x)^2-2`

` = [(sinx+cosecx)^2-2]^2-2` `(because sinx+cosecx = 2)`

` = (4-2)^2-2 = 2`

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

Q 2431356222

If `x = r sin theta cos phi, y = r sin theta sin phi` and `z = r cos theta`,
then `x^2 + y^2 + z^2` is independent of which of the
following?
NDA Paper 1 2007

(A)

Only r

(B)

`r , phi`

(C)

`theta , phi`

(D)

`r , theta`

Solution:

Here, `x = r sin theta cos phi, y = r sin theta sin phi`
and `z = rcos theta`

Now `x^2+y^2+z^2 = r^2 sin^2 theta cos^2 phi+r^2 sin^2 phi sin^2 theta+ r^2 cos^2 theta`

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

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

Thus `x^2+y^2+z^2` is independent of `theta` and `phi`

Correct Answer is `=>` (C) `theta , phi`

Q 2471156026

If `x =a sec theta cos phi , y = b sec theta sin phi` and `z = c tan theta`,

then what is `x^2/a^2 + y^2/b^2-z^2/c^2`
equal to?
NDA Paper 1 2007

(A)

`1`

(B)

`0`

(C)

`-1`

(D)

`a^2+b^2-c^2`

Solution:

Given that, `x = a sec theta cos phi , y = b sec theta sin phi`,
and `z = c tan theta`

`x^2/a^2+y^2/b^2-z^2/c^2 = (a^2 sec^2 theta cos^2 phi)/a^2+(b^2 sec^2 theta sin^2 phi)/b^2 -(c^2 tan^2 theta)/c^2`

` = sec^2 theta(cos^2 phi+ sin^2 phi)-tan^2 theta`

` = sec^2 theta-tan^2 theta = 1`

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

Q 2401545428

What is the value of `(sin 22 1^0/2+cos22 1^0/2)^4 ?`

NDA Paper 1 2008

(A)

`(3+2sqrt2)/2`

(B)

`(1+2 sqrt2)/2`

(C)

`(3sqrt2+2)/2`

(D)

`1`

Solution:

`(sin 22 1^0/2+ cos 22 1^0/2)^4`

` = [(sin^2 22 1^0/2 + cos 22^2 \ 1/2^0) +2 sin 22 1^0/2 cos 22 1^0/2]^2`

` = (1+sin45^0)^2` `(because sin^2 theta+cos^2 theta = 1` and `sin2 theta = 2sintheta * costheta)`

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

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

`= (3+2sqrt2)/3`

Correct Answer is `=>` (A) `(3+2sqrt2)/2`

Q 2421334221

`x = sin theta cos theta` and `y =sin theta + cos theta` are satisfied by
which one of the following equations?
NDA Paper 1 2009

(A)

`y^2-2x = 1`

(B)

`y^2+2x = 1`

(C)

`y^2-2x = -1`

(D)

`y^2+2x = -1`

Solution:

Here, `x = sin theta cos theta` and `y = sin theta + cos theta`

`therefore y^2 - 2x = (sin theta + cos theta)^2 - 2 sin theta cos theta`

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

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

Correct Answer is `=>` (A) `y^2-2x = 1`

Q 2441001823

If `sin x + cosec x = 2`, then what· is the value of
`sin^4 x + cosec^4 x ?`
NDA Paper 1 2009

(A)

`2`

(B)

`4`

(C)

`8`

(D)

`16`

Solution:

Given, `sin x +cosec x = 2`

`therefore sin^4x+cosec^4x = (sin^2x+cosec^2 x)^2-2`

` = [(sinx+cosecx)^2-2]^2-2` `(because sinx+cosecx = 2)`

` = (4-2)^2-2 = 2`

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

Q 2460878715

If `tan theta = sqrtm`, where `m` is a non-square natural
number `m`, then `sec 2theta` is
NDA Paper 1 2011

(A)

a negative number

(B)

a transcendental number

(C)

an irrational number

(D)

a rational number

Solution:

Given, `tan theta = sqrtm` ..............(i)

`sec2theta = sqrt(1+tan^2 2theta) = sqrt(1+ ((2tantheta)/(1-tan^2 theta))^2`

` = sqrt(1+{(2sqrtm)/(1-m)}^2` [from Eq. (i)]

` = sqrt((1-m)^2+4m)/(|(1-m)|) `

` = sqrt(1^2+m^2-2m+4m)/(|(1-m)|)`

` = sqrt((1+m)^2)/(|(1-m)|)`

` = (1+m)/(|(1-m)|)`
Hence, a rational number for every non-square natural number m.

Correct Answer is `=>` (D) a rational number

Q 2210591419

The value of `sin ^2 5^0 + sin^ 2 10^0 +sin ^2 15^0 +sin^ 2 20^0
+ .......... + sin ^2 90^0` is
NDA Paper 1 2015

(A)

`7`

(B)

`8`

(C)

`9`

(D)

`(19)/2`

Solution:

`sin ^2 5^0 + sin^ 2 10^0 +sin ^2 15^0 +sin^ 2 20^0 + ........... + sin ^2 90^0`

`= (sin^2 5^0+ sin^2 85^0) + (sin^2 10^0 + sin^2 80^0)+ ... +sin^2 45^0 + ......+ sin ^2 90^0`

`= sin^2 5^0 + sin^ 2 (90^0 - 5^0)+ sin^ 2 10^0 + sin ^2 (90^0 -10^0)+ ......... + sin^ 2 90^0`

`=(sin^ 2 5^0+ cos^2 5^0)+ (sin ^2 10^0 + cos^ 2 10^0) + .......... + sin ^2 90^0`

`=1+1+1+1+1+1+1+1+(1/sqrt(2))^2 +1`

`= 9 + 1/2 = (19)/2`

Correct Answer is `=>` (D) `(19)/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`

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

Q 2400556418

Which one of the following is positive in the third
quadrant?
NDA Paper 1 2012

(A)

`sintheta`

(B)

`cos theta`

(C)

`tantheta`

(D)

`sec theta`

Solution:

The conditions of trigonometric function in different
quadrant as follows.

Hence, `tan theta` lies in third quadrant.

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

Q 2430623512

What is `(1 - sin^2 theta) (1 + tan^2 theta)` equal to'?
NDA Paper 1 2013

(A)

`sin^2 theta`

(B)

`cos^2 theta`

(C)

`tan^2 theta`

(D)

`1`

Solution:

`(1 - sin^2 theta) (1 + tan^2 theta) = cos^2 theta * sec^2 theta`

` = cos^2 theta * 1/(cos^2 theta) = 1`

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

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 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 2440556413

What is the value of `sin A cos A tan A + cos A sinA cotA?`
NDA Paper 1 2012

(A)

`sin A`

(B)

`cos A`

(C)

`tan A`

(D)

`1`

Solution:

Let `f(A) =sin A ·cosA ·tan A+ cosA·sinA·cotA`

` = sinA * cos A * (sinA)/(cosA) + cosA * sinA * (cosA)/(sinA)`

` = sin^2A+cos^2A = 1` `(because sin^2 theta+cos^2 theta = 1) .`

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

Q 2400656518

What is the value of `(sintheta)/(cosectheta)+(costheta)/(sectheta) ?`

NDA Paper 1 2012

(A)

`1`

(B)

`1/2`

(C)

`1/3`

(D)

`2`

Solution:

given `(sintheta)/(cosectheta)+(costheta)/(sectheta) `

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

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

Sum and Difference of Two Angles and Transformation Formulae

Sum and Difference of Two Angles
(i) `sin (A + B) = sin A cos B + cos A sin B \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ sin (A - B) = sin A cos B - cos A sin B`

(ii) `cos (A + B)= cos A cos B - sin A sin B \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ cos (A -B)= cos A cos B + sin A sin B`

(iii) `tan (A + B) =(tan A + tan B)/(1 - tan A tan B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \tan (A - B) =(tan A - tan B)/(1 + tan A tan B) `

(iv) `cot (A + B) = (cot A cot B - 1)/( cot B + cot A)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \cot (A - B) = (cot A cot B + 1)/( cot B - cot A)`

(v) `tan(A+B+C) = (tanA+tanB+tanC-tanAtanBtanC)/(1-tanAtanB -tanBtanC-tanCtanA) `

(vi) `sin (A+ B) sin (A-B)= sin^2 A - sin^2 B = cos^2 A - cos^2 A`

(vii) `cos(A+B) cos(A-B) = cos^2 A - sin^2 B = cos^2 B - sin^2 A`

Transformation Formulae
(i) `2sinA cos B =sin (A+ B)+ sin (A -B)`

(ii) `2cos A sin B = sin (A +B)- sin (A -B)`

(iii) `2cosA cosB=cos(A +B)+cos(A -B)`

(iv) `2sinAsinB=cos(A+ B)-cos(A-B)`

(v) `sin C+ sin D = 2 sin ((C+D)/2) cos((C-D)/2)`

(vi) `sin C- sin D = 2 sin ((C-D)/2) cos ((C+D)/2)`

(vii) `cos C+ cos D = 2 cos((C+D)/2) cos((C-D)/2)`

(viii) `cos C- cos D= 2 sin ((C+D)/2) sin ((D-C)/2)`

Q 2763680545

If `K = sin (pi/18) sin((5pi)/18) sin((7pi)/18)` then what is the value of K?
NDA Paper 1 2017

(A)

`1/2`

(B)

`1/4`

(C)

`1/8`

(D)

`1/16`

Solution:

`K = sin (pi/18) sin((5pi)/18) sin((7pi)/18)`

`sin \ \ pi/18 sin \ \ (5pi )/18 sin \ \(7 pi )/18 = sin 10° sin 50 ° sin 70 °`

`= 1/2 sin 10 ° [ 2 sin 70 ° sin 50° ]`

`= 1/2 sin 10° [ cos (70 -50 ) - cos ( 70 + 50 ) ]`

`= 1/2 sin 10 ° [ cos 20° - cos 120° ]`

`= 1/2 sin 10 ° [ 1 - 2 sin^2 10° - ( -1/2 ) ] = 1/2 sin 10° [ 3/2 - 2 sin^2 10° ]`

`= 1/2 sin 10° [ ( 3 - 4 sin^2 10° ) /2 ] = 1/4 [ 3 sin 10° - 4 sin ^3 10 ° ]`

where ` sin 3 theta = 3 sin theta - 4 sin ^3 theta `

`:. sin 3 theta = sin 3 xx 10 ° = sin 30° = 1/2`

So ` K = 1 /4 xx 1/2 = 1/8`

Correct Answer is `=>` (C) `1/8`

Q 2783680547

The expression `( sin alpha + sin beta)/( cos alpha + cos beta )` is equal to
NDA Paper 1 2017

(A)

`tan ((alpha+beta)/2)`

(B)

`cot ((alpha+beta)/2)`

(C)

`sin ((alpha+beta)/2)`

(D)

`cos ((alpha+beta)/2)`

Solution:

`( sin alpha + sin beta)/( cos alpha + cos beta )`

`= ( 2 sin \ \( alpha + beta/2 ) cos \ \((alpha - beta ) )/2 )/( 2 cos ( (alpha + beta )/ 2 ) cos (( alpha - beta )/2 ) )`

` = tan ( ( alpha + beta ) /2 )`

Correct Answer is `=>` (A) `tan ((alpha+beta)/2)`

Q 2713780640

If `sintheta= 3sin(theta +2alpha)`, then the value of `tan (theta +alpha) +2 tanalpha` is equal to
NDA Paper 1 2017

(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

`(sin(theta+ 2 alpha))/(sin theta) = 1/3`

`(sin (theta+ 2 alpha) + sin theta)/(sin (theta+ 2 alpha) - sin theta) = 4/(-2) =-2`

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

`tan (alpha+ theta) + 2 tan alpha=0`

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

Q 2117501489

If `A = (cos 12^0 - cos 36^0 ) (sin 96^0 + sin 24^0)` and
`B = (sin 60^0 - sin 12^0 ) (cos 48^0 - cos 72^0 )`, then what is
`A/B` equal to?
NDA Paper 1 2016

(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

We have, `A = (cos 12^0 - cos 36^0) (sin 96^0 + sin 24^0 )`

and `B = (sin 60° - sin 12°) (cos 48^0 - cos 72°)`

Now, consider

`A = (cos 12°- cos 36°) (sin 96^0 + sin 24°)`

`= 2 sin 24° sin ( -12°) (2 sin 60° cos 36°)`

`= 4 sin 24° sin 12° sin 60° cos 36°`

and `B = 2 cos 36° sin 24° (-2 sin 60° sin (-12°))`

`= 4 sin 24° sin 12° sin 60^0 cos 36°`

Clearly, `A/B = 1`

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

Q 2400145018

The expression `(cotx+cosec-1)/(cotx-cosec+1)` is equal to
NDA Paper 1 2013

(A)

`(sinx)/(1-cosx)`

(B)

`(1-cosx)/(sinx)`

(C)

`(1+cosx)/(sinx)`

(D)

`(sinx)/(1+cosx)`

Solution:

`(cotx+cosec-1)/(cotx-cosec+1)`

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

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

` = (2 cos (x/2) (cos (x/2)- sin (x/2)))/(2 sin (x/2)(cos (x/2)-sin (x/2))) = (2 cos (x/2))/(2 sin (x/2)) = (2cos (x/2) * cos (x/2))/(2 sin (x/2) * cos (x/2))`

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

Correct Answer is `=>` (C) `(1+cosx)/(sinx)`

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 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 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 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 2460891715

If `tan A =3/4 ` and `tan B = -12/5` , then how many

values can `cot (A - B)` have depending on the
actual values of `A` and `B`'?
NDA Paper 1 2010

(A)

`1`

(B)

`2`

(C)

`3`

(D)

`4`

Solution:

`because tanA = 3/4 ` and `tanB = -12/5`

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

which shows that `cot (A - B)` has only two value of `A` and `B`.

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

Q 2431823722

What is the value of `1- sin 1 0^0 sin 50^0 sin 70^0?`
NDA Paper 1 2009

(A)

`1/8`

(B)

`3/8`

(C)

`5/8`

(D)

`7/8`

Solution:

`1 - sin 10^0 sin 50^0 sin 70^0`

` = 1-1/2(2sin70^0 sin10^0) sin50^0`

`1-1/2[(cos60^0-cos80^0)sin50^0]`

` = 1-1/2(1/2 sin50^0-cos80^0sin50^0)`

` = 1-1/4(sin50^0-2cos80^0 sin50^0)`

` = 1-1/4[(sin50^0-sin130^0)+sin30^0]`

` = 1-1/4(-2cos90^0*sin40^0+sin30^0)`

` = 1-1/4(0+1/2) = 1-1/8 = 7/8`

Correct Answer is `=>` (D) `7/8`

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 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 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 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 2440180013

If `tan A = 1/2` and `tan B = 1/3,` then what is the
value of `4A + 4B?`
NDA Paper 1 2011

(A)

`pi/4`

(B)

`pi/2`

(C)

`pi`

(D)

`2pi`

Solution:

Given, `tan A = 1/2` and `tanB = 1/3`

`=> tan(A+B) = (tanA+tanB)/(1-tanA*tanB) = (1/2+1/3)/(1-1/6) = (5/6)/(5/6)`

` = 1 = tan(pi/4)`

`=> A+B = pi/4` ...............(i)

`=> 4A+4B = pi`

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

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 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 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 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 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 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 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 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`

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 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 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`

Trigonometric Ratios of Multiple of an Angle and Trigonometric Ratios of Submultipies Angles

Trigonometric Ratios of Multiple of an Angle
(i) `sin 2A = 2 sin A cos A =(2 tan A)/(1+ tan^2 A)`

(ii) `cos 2 A = cos^2 A - sin^2 A = 2 cos^2 A-1=1-2 sin^2 A=(1- tan^2 A)/(1+ tan^2 A)`

(iii) `tan 2 A =(2 tan A)/(1- tan^2 A)`

(iv) `cot 2A = (cot^2 A -1)/(2 cot A)`

(v) `sin 3 A = 3 sin A - 4 sin^3 A`

(vi) `cos 3 A = 4 cos^3 A- 3 cos A`

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

(viii) `cot 3 A =(3 cot A - cot ^3 A)/(1- 3 cot^2 A)`

Trigonometric Ratios of Sub multiples Angles

(i) `sin A = 2 sin \ (1/2 A) cos \ (1/2 A) = ( 2 tan \ (1/2 A))/(1+ tan^2 \ (1/2 A))`

(ii) `cos A = cos^2 \ (1/2 A) - sin^2 \ (A/2) = 2 cos^2 \ (1/2 A)-1= 1- 2 sin^2 \ (1/2 A) = (1- tan^2 \ (1/2 A))/(1+ tan^2 \ (1/2 A))`

(iii) `tan A = ( 2 tan \ (1/2 A))/(1- tan^2 \ (1/2 A))`

(iv) `cot A = (cot^2 \ (1/2 A)-1)/(2 cot \ (1/2 A))`

Q 2713880740

If `tan (alpha + beta) = 2` and `tan (alpha - beta) = 1`, then `tan (2alpha)` is equal to
NDA Paper 1 2017

(A)

`-3`

(B)

`-2`

(C)

`-1/3`

(D)

`1`

Solution:

`tan (alpha+beta)=2 \ \ \ \ \ tan(alpha-beta)=1`

`tan 2 alpha = tan [(alpha+ beta)+ ( alpha-beta)]`

`=(tan (alpha+ beta) + tan (alpha- beta)]/(1- tan (alpha+ beta) tan (alpha- beta)]`

`=-3`

Correct Answer is `=>` (A) `-3`

Q 2410445310

What is the value of `(1-tan^2 (x/2))/(1+tan^2 (x/2) `?

NDA Paper 1 2013

(A)

`sinx * cos x`

(B)

`tanx`

(C)

`sinx`

(D)

`cosx`

Solution:

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

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

` = cosx`

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

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 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 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 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 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 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 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 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 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 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 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 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 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`