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Tricks & Tips of Basic Trigonometry FOR NDA PART - 1

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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`

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 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)`

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