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Tricks & Tips Of Inverse Trigonometric Functions For NDA

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Finding Principal value of ITF

1. `y = sinx `
Domain : `-pi/2 le x le pi/2`
Range : `-1 le y le 1`

`=> y = sin^(-1)x`
Domain : ` -1 le x le 1`
Range : `-pi/2 le y le pi/2`


2. `y = cosx`
Domain : `0 le x le pi`
Range : `-1 le y le 1`

`=> y = cos^(-1)x`
Domain : ` -1 le x le 1`
Range : `0 le y le pi`


3. `y=tan x`
Domain : `-pi/2 < x < pi/2`
Range : `-oo < y < oo`

`=> y = tan^(-1)x`
Domain : ` -oo < x < oo`
Range : `-pi/2 < y < pi/2`


4. `y= cosec (x)`
Domain : `-pi/2 le x le pi/2 , x ne 0`
Range : `y le -1` or `y ge 1`

`=> y = cosec^(-1)x`
Domain : ` x le -1` or `x ge 1`
Range : `-pi/2 le y le pi/2 , y ne 0`


5. `y= sec (x)`
Domain : `0 le x le pi , x ne pi/2`
Range : `y le -1` or `y ge 1`

`=> y = sec^(-1)x`
Domain : ` x le -1` or `x ge 1`
Range : `0 le y le pi , y ne pi/2`


6. `y= cot (x)`
Domain :`0 < x < pi`
Range : `-oo < y < oo`

`=> y = sec^(-1)x`
Domain : ` oo < x < oo`
Range : `0 < y< pi`
Q 2460145015

What is the principle value of `sec^(-1) (2/sqrt 3)` ?
NDA Paper 1 2011
(A)

`pi/2`

(B)

`pi/3`

(C)

`pi/4`

(D)

`pi/6`

Solution:

Given, `sec^(-1) (2/sqrt 3)`

` ∵` Range of `sec^(-1) x=[0, pi/2) cup (pi/2,pi]=sec^(-1)[sec(pi/6)]=pi/6`

which is the required principle value.
Correct Answer is `=>` (D) `pi/6`
Q 2480745617

What is the principle value of `cosec^(-1) (- sqrt 2)` ?
NDA Paper 1 2010
(A)

`pi/4`

(B)

`pi/2`

(C)

`-pi/4`

(D)

`0`

Solution:

Principle value of `cosec^(-1) (-sqrt 2)=-pi/4`

So, the range of `cosec^(-1) x` is `[-pi/2,0) cup (pi/2 ,pi]`
Correct Answer is `=>` (C) `-pi/4`
Q 2480245117

Consider the following statements

I. `cosec^(-1) (-2/sqrt 3)=-pi/3`

II. `sec^(-1) (2/sqrt 3)=pi/6`

Which of the above statements is/are correct?
NDA Paper 1 2011
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

`cosec^(-1) (-2/sqrt 3)=-pi/3`

` ∵` The range of `cosec^(-1) x` is `[-pi/2 ,0) cup (pi/2, pi]`

and `sec^(-1) (2/sqrt 3)=pi/6`

Since, the range of `sec^(-1) x ` is `[0, pi/2) cup (pi/2, pi]`

Hence, both Statements I and II are correct
Correct Answer is `=>` (C) Both I and II
Q 2675378266

Find the principal value of `cos^(–1)x`, for `x= sqrt 3/2`

Solution:

If `cos ^(-1) (sqrt 3/2) = theta` , then `cos theta = sqrt 3/2`

Since we are considering principal branch, `θ ∈ [0, π]`. Also, since `sqrt 3/2 > 0` `θ` being in

the first quadrant, hence `cos^(–1) (sqrt 3/2) = pi/6`
Q 2130645512

Find the principal values of `sec^(-1) (2/sqrt3)`
Class 12 Exercise 2.1 Q.No. 7
Solution:

Let `sec^(-1) (2/sqrt3) =y`

`=> sec y =2/sqrt3 = sec pi/6`

The range of principal value of `sec^(-1)` is

`[0, pi] - (pi/2)`


`:.` The principal value of `sec^(-1) (2/sqrt3)` is `pi/6`

Simplification using triangular property and Property 1

`sin (sin^-1 x) = x, x ∈ [– 1, 1]` and `sin^-1 (sin x) = x, x ∈ [ - pi/2, pi/2]`

`cos (cos^-1 x) = x, x ∈ [– 1, 1]` and `cos ^-1 (cos x) = x, x ∈ [ 0,pi]`

`tan (tan^-1 x) = x, x ∈ R` and `tan^-1 (tan x) = x, x ∈ ( - pi/2, pi/2)`

`cosec (cosec^-1 x) = x, |x| ge 1` and `cosec^-1 (cosec x) = x, x ∈ [ - pi/2, pi/2] , xne0`

`sec (sec^-1 x) = x, |x| ge 1` and `sec^-1 (sec x) = x, x ∈ [ 0, pi], x ne pi/2`

`cot (cot^-1 x) = x, x ∈ R` and `cot^-1 (cot x) = x, x ∈ ( 0,pi)`

Q 2460656515

What is the value of

`cos[tan^(-1){tan \ (15 pi)/4}]` ?
NDA Paper 1 2007
(A)

`-1/sqrt 2`

(B)

`0`

(C)

`1/sqrt 2`

(D)

`1/(2 sqrt 2)`

Solution:

`cos [tan((15 pi)/4)}]`

`=cos[tan^(-1){tan(4 pi-pi/4)}]`

`=cos[tan^(-1){tan (-pi/4)}]=cos(-pi/4)`

`=cos(pi/4)= +1/sqrt 2`
Correct Answer is `=>` (C) `1/sqrt 2`
Q 2410656510

If `sin^(-1) x = tan^(-1) y`, then what is the value of `1/x^2-1/y^2` ?
NDA Paper 1 2007
(A)

`1`

(B)

`-1`

(C)

`0`

(D)

`2`

Solution:

Given that, `sin^( -1) x = tan^(-1) y`

We know that, `tan^(-1) y = sin^(-1)(y/(sqrt(1+y^2)))`


`:. sin^(-1) (x)=sin^(-1) (y/(sqrt(1+y^2)))=> x=y/(sqrt(1+y^2))`

`=> x^2(1+y^2)=y^2=>(1+y^2)/y^2=1/x^2=> 1/y^2+1=1/x^2`

`:. 1/x^2-1/y^2=1`
Correct Answer is `=>` (A) `1`
Q 2400456318

The equation `sin^(-1) (3x -4x^3) =3sin^(- 1) (x)` is true
for all values of `x` lying in which one of the
following intervals?
NDA Paper 1 2008
(A)

`[-1/2,1/2]`

(B)

`[1/2,1]`

(C)

`[-1,-1/2]`

(D)

`[-1,1]`

Solution:

The equation `sin^(-1) (3x - 4x^3 ) = 3 sin^(- 1) x` is true for all
values of `x` lying in the interval `[-1/2,1/2]` (by property)
Correct Answer is `=>` (A) `[-1/2,1/2]`
Q 2569691515

The value of `cos { tan^(-1) ( tan (15 pi)/4 ) }` is
BCECE Mains 2015
(A)

` 1/sqrt2`

(B)

` - 1/sqrt2`

(C)

`1`

(D)

None of these

Solution:

We have, `cos { tan^(-1) ( tan (15 pi)/4 ) }`

` = cos { tan^(-1) ( tan (4 pi - pi/4 )) }`

` = cos { tan^(-1) tan (- pi//4)}`

`= cos (-pi//4)`

`= cos pi//4 = 1//sqrt2`
Correct Answer is `=>` (A) ` 1/sqrt2`
Q 2460356215

What is `tan (cos^(-1) x)` equal to?
NDA Paper 1 2008
(A)

`(sqrt(1-x^2))/x`

(B)

`x^2/(1+x^2)`

(C)

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

(D)

`sqrt(1-x^2)`

Solution:

Let `cos^(-1) x=theta`

`=>cos theta=x`

`:. tan theta=(sqrt(1-x^2))/x`

Hence, `tan (cos^(-1)x)=(sqrt(1-x^2))/x`
Correct Answer is `=>` (A) `(sqrt(1-x^2))/x`
Q 2480256117

Which one of the following is not correct?
NDA Paper 1 2008
(A)

`sin^(-1) {sin (5pi //4)} = - pi // 4`

(B)

`sec^(-1) {sec (5pi //4)} = 3pi // 4`

(C)

`tan^(-1) {tan (5pi //4)} = pi// 4`

(D)

`cosec^(-1) {cosec (7pi//4)} = pi//4`

Solution:

(a) `sin^(-1) {sin 5(pi /4)} = - pi / 4=sin^(-1) {sin(pi+pi/4)}`

`=sin^(-1){-sin (pi/4)}=-pi/4` (correct)

(b) `sec^(-1) {sec 5(pi /4)} = 3pi / 4=sec^(-1){sec(2pi-(3 pi)/4)}`

`=sec^(-1){sec3(pi/4)}=(3 pi)/4` (correct)

(c) `tan^(-1) {tan (5pi /4)} = pi/4= tan^(-1){tan(pi+pi/4)}`

`=tan^(-1){tan (pi/4)}=pi/4` (correct)

(d) `cosec^(-1) {cosec (7pi/4)} = pi/4=cosec^(-1){cosec(2 pi-pi/4)}`

`=cosec^(-1){-cosec (pi/4)}=-pi/4` (incorrect)
Correct Answer is `=>` (D) `cosec^(-1) {cosec (7pi//4)} = pi//4`
Q 2430256112

What is the value of `sin [ sin^(-1) (3/5)+sin^(-1) (4/5)]` ?
NDA Paper 1 2012
(A)

`0`

(B)

`1//2`

(C)

`1`

(D)

`2`

Solution:

`sin[sin^(-1) (3/5)+sin^(-1) (4/5)]`

Let `sin^(-1)(4/5)=theta=> sin theta=4/5`

Now, `cos theta=sqrt(1-sin^2 theta)=sqrt(1-(4/5)^2)`

`=sqrt(1-16/25)=sqrt(9/25)=3/5`

`=> cos theta =3/5=> theta=cos^(-1)(3/5)`

`:. sin[sin^(-1)(3/5)+sin^(-1)(4/5)]=sin[sin^(-1)(3/5)+cos^(-1)(3/5)]`

`( ∵ sin^(-1)+cos^(-1) x=pi/2)`

`=sin (pi/2)=1`
Correct Answer is `=>` (C) `1`
Q 2420145911

Statement I : If `-1 le x < 0`, then

`cos (sin^(-1) x) =-sqrt(1-x^2)`

Statement II : If `-1 le x < 0`, then

`sin (cos^(-1)=sqrt(1-x^2))`

Which one of the following is correct in respect of the
above statements?
NDA Paper 1 2010
(A)

Both Statements I and II are independently correct and Statement II is the correct explanation of Statement I

(B)

Both Statements I and II are independently correct but Statement II is not the correct explanation of Statement I

(C)

Statement I is correct but Statement II is false

(D)

Statement I is false but Statement II is correct

Solution:

I. `cos (sin^(-1 )x) =cos (cos^(-1) sqrt (1- x^ 2) ) = sqrt(1- x^ 2)` ,

when `x in [-1, 1]`

II. `sin (cos^(-1 ) x) =sin (sin^(-1) sqrt (1-x^2))`, when `x in [-1. 1`]

`= sqrt(1- x^2)`

Hence, Statement I is false and Statemet II is true.
Correct Answer is `=>` (D) Statement I is false but Statement II is correct
Q 1732645532

What is `sin^(-1) sin ((3pi)/5)` equal to?
NDA Paper 1 2014
(A)

`(3pi)/5`

(B)

`(2 pi)/5`

(C)

`( pi)/5`

(D)

None of these

Solution:

We have,

`sin^(-1) sin ((3pi)/5)`

` = sin^(-1) sin ( pi/2 + pi /(10) )`

` [ ∵ sin^(-1) sin x =x , AA x in ( - pi/2 , pi/2 ) ]`

` = sin^(-1) cos (pi/(10))`

` = sin^(-1) sin x ( pi/2 - pi/(10) )`

` [ ∵ cos theta = sin (pi/2 - theta ) ]`

` = pi/2 = pi/(10)`

` = (5 pi - pi)/(10) = (4pi) /(10) = (2pi)/5`
Correct Answer is `=>` (B) `(2 pi)/5`

Problems based on Property 2

• `sin ^-1 \ \1/x = cosec^-1 x , x >= 1 ` or ` x <= -1`

• `cos^-1 \ \ 1/x = sec^-1 x, x ≥ 1` or `x ≤ – 1`

• `tan ^-1 \ \ 1/x = cot^-1 x , x > 0`

To prove the first result, we put `cosec^-1 x = y`, i.e., x = cosec y

Therefore `1/x = sin y`

Hence `sin^-1 \ \1/x = y`

or ` sin ^-1 \ \1/x = cosec^-1 x`

Similarly, we can prove the other parts.


p3

(i) `sin^-1 (–x) = – sin^-1 x, x ∈ [– 1, 1]`

(ii) `tan^-1 (–x) = – tan^-1 x, x ∈ R`

(iii) `cosec^-1 (–x) = – cosec^-1 x, | x | ≥ 1`

Let `sin^-1 (–x) = y`, i.e., `–x = sin y` so that `x = – sin y`, i.e., `x = sin (–y)`.

Hence `sin^-1 x = – y = – sin^-1 (–x)`

Therefore `sin^-1 (–x) = – sin^-1 x`

Similarly, we can prove the other parts.

(i) `cos^-1 (–x) = π – cos^-1 x, x ∈ [– 1, 1]`

(ii) `sec^-1 (–x) = π – sec^-1 x, | x | ≥ 1`

(iii) `cot^-1 (–x) = π – cot^-1 x, x ∈ R`

Let `cos^-1 (–x) = y` i.e., `– x = cos y` so that `x = – cos y = cos (π – y)`

Therefore `cos^-1 x = π – y = π – cos^-1 (–x)`

Hence `cos^-1 (–x) = π – cos^-1 x`

Similarly, we can prove the other parts.
Q 2410756610

If `cos^(-1)(1/sqrt 5)=theta`, then what is the value of
`cosec^(-1) (sqrt 5)` ?
NDA Paper 1 2007
(A)

`pi/2+theta`

(B)

`pi/2-theta`

(C)

`pi/2`

(D)

`-theta`

Solution:

`cos^(-1) (1/sqrt 5)=theta`

`=>sec^(-1) (sqrt 5)=theta` `[ cos^(-1)(1/x)=sec^(-1)x]`

`=> pi/2-cosec^(-1) (sqrt 5)=theta` `(∵ sec^(-1)x+cosec^(-1)x=pi/2)`

`=> cosec^(-1)(sqrt 5)=pi/2-theta`
Correct Answer is `=>` (B) `pi/2-theta`

Problems based on Property 3

(i) `sin^-1 x + cos^-1 x = π/2, x ∈ [– 1, 1]`

(ii) `tan^-1 x + cot^-1 x = π/2, x ∈ R`

(iii) `cosec^-1 x + sec^-1 x = pi/2 , | x| >= 1`

Let `sin^-1 x = y`. Then `x = sin y = cos ( pi/2 - y)`

Therefore `cos^-1 x = pi/2 - y = pi/2 - sin ^-1 x`

Hence `sin^-1 x + cos^-1 x = pi/2`

Similarly, we can prove the other parts.
Q 2420456311

If `sin ^(-1) x - cos ^(-1) x = pi/6` , then what is the value of `x`?
NDA Paper 1 2008
(A)

`x=-1/2`

(B)

`x=1`

(C)

`x=1/2`

(D)

`x=sqrt 3/2`

Solution:

Given that, `sin^(-1) x- cos^(-1) x = pi/6`................(i)

We know that `sin^(-1)x+cos^(-1)x=pi/2`..............(ii)

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

`2 sin^(-1)x=pi/6+pi/2=(2 pi)/3=> sin^(-1) x=pi/3`

`:. x=sin(pi/3)=sqrt 3/2`
Correct Answer is `=>` (D) `x=sqrt 3/2`
Q 2430356212

If `sin^(- 1) x+sin^(- 1) y=pi/2` and `cos^(-1)x-cos^(-1)y=0`,

then the values `x` and `y` are respectively
NDA Paper 1 2008
(A)

`1/sqrt 2,-1/sqrt 2`

(B)

`1/2, 1/2`

(C)

`1/2, -1/2`

(D)

`1/sqrt 2,1/sqrt 2`

Solution:

Given, `sin^(-1) x + sin^(-1) y = pi/2`..............(i)

and `cos^(-1 ) x- cos^(-1) y = 0`

`=>(pi/2-sin^(-1) x)-(pi/2-sin^(-1) y)=0`

`=>sin^(-1) y- sin^(-1) x = 0`

`=> sin^(-1) y = sin^(- 1) x` ....... (ii)

From Eqs. (i) and (ii),

`2 sin^(-1) x =pi/2=>sin ^(-1)x= pi/4`

`=> x= sin \ pi/4 => x = 1/sqrt2`

From Eq. (ii),

`x= 1/sqrt 2`
Correct Answer is `=>` (D) `1/sqrt 2,1/sqrt 2`
Q 2450845714

If `sin^(-1) (5/x)+sin^(-1)(12/x)=pi/2`, then what is the value of `x` ?
NDA Paper 1 2010
(A)

`1`

(B)

`7`

(C)

`13`

(D)

`17`

Solution:

`sin^(-1) (5/x)+sin^(-1) (12/x)=pi/2`

`=> sin^(-1) (5/x)+cos^(-1) \ (sqrt(x^2-144))/x =pi/2`

But `sin^(-1) y+cos^(-1) y=pi/2`

`:. 5/x= (sqrt(x-144))/x`

`=>25=x^2-144=> x^2=169`

`:. x=13`
Correct Answer is `=>` (C) `13`
Q 2610812719

If `tan^(-1)\ \ a/x + tan^(-1)\ \ b/x = pi//2`, then `x` is equal to
BCECE Mains 2015
(A)

` sqrt(ab)`

(B)

` sqrt(2ab)`

(C)

`2ab`

(D)

`ab`

Solution:

We have,

` tan^(-1) \ \ a/x + tan^(-1) \ \ b/x = pi/2`

` => tan^(-1) ( (a/x + b/x)/(1 - (ab)/x^2) ) = pi/2`

` => 1 - (ab)/x^2 = 0`

` => x = sqrt(ab)`
Correct Answer is `=>` (A) ` sqrt(ab)`
Q 1523080841

If `4\sin^{-1}x+\cos^{-1}x=\pi`, then `x` is equal to:
BITSAT 2012
(A)

`\frac{1}{2}`

(B)

`2`

(C)

`1`

(D)

`\frac{1}{3}`

Solution:

`4\sin^{-1}x+\cos^{-1}x=\pi`

`\Rightarrow 3\sin^{-1}x+(\sin^{-1}x+\cos^{-1}x)=\pi`

`\Rightarrow 3\sin^{-1}x+\ frac{\pi}{2}=\pi`

`\Rightarrow \sin^{-1}x=\ frac{\pi}{6}`

`\Rightarrow x=\sin\ frac{\pi}{6}=\ frac{1}{2}`
Correct Answer is `=>` (A) `\frac{1}{2}`
Q 2440145913

If `sin^(-1) x+cot^(-1) (1//2)=pi/2`, then what is the
value of `x`?
NDA Paper 1 2009
(A)

`0`

(B)

`1/sqrt 5`

(C)

`2/sqrt 5`

(D)

`sqrt 3/2`

Solution:

`sin^(-1) x+cot^(-1) (1/2)=pi/2`

` ∵ cot^(-1)x=cos^(-1) (x/(sqrt(1+x^2)))`

`=> sin^(-1)x+cos^(-1) (1/sqrt 5)=pi/2` `[ ∵ sin^(-1)theta+cos^(-1) theta=pi/2]`

`:. x=1/sqrt 5`
Correct Answer is `=>` (B) `1/sqrt 5`
Q 2430545412

If `sin [sin^(-1) (1/5) cos^(-1) x]=1` then what is `x` equal to ?
NDA Paper 1 2011
(A)

`0`

(B)

`1`

(C)

`4//5`

(D)

`1//5`

Solution:

`sin [ sin^(-1) (1/5)+ cos^(-1) x]=1`

`=> sin[sin^(-1) (1/5 +cos^(-1) (x)]=sin (pi/2)`

`=> sin^(-1) (1/5)+cos^(-1) x=pi/2`

`=> cos^(-1) x=pi/2-sin^(-1) (1/5)=cos^(-1) (1/5)`

`( ∵ sin^(-1) x+cos^(-1) x =pi/2)`

`:. x=1/5`
Correct Answer is `=>` (D) `1//5`
Q 2166691575

Consider the following statements
1. `tan^(-1) x + tan ^(-1) (1/x) = pi`

2. Their exist, `x, y in [ -1, 1],` where `x != y` such that
` sin^(-1) x + cos^(-1) y = (pi)/2`

Which of the above statements is/are correct?
NDA Paper 1 2016
(A)

Only `1`

(B)

Only `2`

(C)

Both `1` and `2`

(D)

Neither `1` nor `2`

Solution:

We know that, `tan^(-1) x + cot^(-1) x = pi/2`

`:. tan^(-1) x + tan^(-1) (1/x) = pi/2`

`:.` Statement 1 is incorrect.

Let `x=y`

Given that

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

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

This is true for `x in [-1, 1]`.

`:.` Statement `2` is not correct.
Correct Answer is `=>` (D) Neither `1` nor `2`
Q 1772445336

Consider the following statements

I. `tan^(-1) 1 + tan^(- 1) (0.5) = pi//2`

II. `sin ^(-1) (1// 3) +cos^(- 1) (1//3) = pi// 2`

Which of the above statements is//are correct?
NDA Paper 1 2014
(A)

Only 1

(B)

Only 2

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

I. `tan^(-1) 1 + tan^(- 1) (0.5)`

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

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

`( ∵ tan^(-1) x = cot^(- 1) (1/x) )`

` = tan^(-1) + cot^(- 1) 2`

` != pi//2`

` [ ∵ tan^(-1) x = cot^(- 1) x = pi//2 , AA x in R]`

II. `sin^(-1) (1/3) + cos ^(-1) (1/3)`

` = pi//2`

` [ ∵ sin^(-1) x + cos ^(-1) x = pi//2 , AA x in R]`
Correct Answer is `=>` (B) Only 2

Problems based on Property 4

(i) `tan^-1x + tan^-1 y = tan^-1 \ \( x +y)/( 1 - xy) , xy < 1`

(ii) `tan^-1x – tan^-1 y = tan^-1 \ \( x - y)/( 1 + xy), xy > -1`

(iii) `2tan^-1x = tan^-1\ \ (2x ) /( 1- x^2) , | x | < 1`

Let `tan^-1 x = θ` and `tan^-1 y = phi`. Then `x = tan θ, y = tan phi`

Now `tan ( theta + pi) = ( tan theta + tan phi )/( 1- tan theta tan phi ) = ( x + y) /( 1- xy)`

This gives ` theta + phi = tan ^-1 \ \ ( x+y) /(1 -xy)`

Hence ` tan ^-1 a + tan ^-1 y = tan^-1 \ \ ( x +y) /(1 -xy)`

In the above result, if we replace y by – y, we get the second result and by replacing y by x, we get the third result.
Q 2703780648

Let `x, y, z` be positive real numbers such that `x, y, z` are in GP and `tan^(-1) x, tan^(-1) y` and `tan^(-1) z` are in AP. Then which one of the following is correct?
NDA Paper 1 2017
(A)

`x = y = z`

(B)

`xz = 1`

(C)

`x ne y` and `y=z`

(D)

`x = y` and `y ne z`

Solution:

`2y = x + z ` … (1)

As `tan^(−1) x, tan^(−1) y, tan^(−1) z` in AP

`⇒2 tan^(−1 ) y = tan^(−1 ) ((x+z)/(1−xz))`

`(2y)/(1−y^2)=(x+z)/(1−xz)`

`(x+z)/(1−y^2)=(x+z)/(1−xz)` .....by (1)

` (x+z){1/(1−y^2) −1/(1−xz)}=0`

` x+z=0` or ` 1−xz=x−y^2x+z=0`

`y^2 =xz`
`⇒ x, y, z` in GP.

As `x, y, z` AP & GP

`⇒ x = y = z`
Correct Answer is `=>` (A) `x = y = z`
Q 1732878732

If `x` and `y` are positive and `xy > 1`, then what is
`tan^( -1) x +tan^( -1) y` equal to?
NDA Paper 1 2014
(A)

`tan^( -1) ( (x + y)/(1 - xy))`

(B)

` - pi + tan^( -1) ( (x + y)/(1 - xy))`

(C)

` pi - tan^( -1) ( (x + y)/(1 - xy))`

(D)

`tan^( -1) ( (x - y)/(1 + xy))`

Solution:

We know that,

If `x` and `y` are positive i.e., `x > 0, y > 0` and `xy > 1`. then

`tan^(-1) x + tan^(-1) y = pi + tan^(-1) ( (x + y)/(1 - xy))`

While, if `x < 0, y < 0` and `xy > 1`, then

`tan^(-1) x +tan^(- 1) y = - pi + tan^(- 1) ( (x + y)/(1 - xy))`
Correct Answer is `=>` (B) ` - pi + tan^( -1) ( (x + y)/(1 - xy))`
Q 2400378218

What is the value of `tan^(-1)(m/n)-tan^(-1) ((m-n)/(m+n))` ?
NDA Paper 1 2007
(A)

`pi`

(B)

`pi/2`

(C)

`pi/4`

(D)

`pi/3`

Solution:

`tan^(-1) (m/n)-tan^(-1) ((m-n)/(m+n))`

`tan^(-1) (m/n)-tan^(-1) ((1-n/m)/(1+n/m))=tan^(-1)(m/n)-tan^(-1) ((1-n/m)/(1+1*n/m))`

`= tan^(-1) (m/n)-{tan^(-1)(1) -tan^(-1) n/m)}`

`=tan^(-1) (m/n)+tan^(-1) (n/m)-pi/4`

`=tan^(-1) (m/n)+cot^(-1) (m/n)-pi/4`

`=pi/2-pi/4=pi/4`
Correct Answer is `=>` (C) `pi/4`
Q 2440156013

If `sin^(-1) ((2a)/(1+a^2))-cos^(-1) ((1-b^2)/(1+b^2))=tan^(-1)((2x)/(1-x^2))`
then what is the value of `x`?
NDA Paper 1 2009
(A)

`a/b`

(B)

`ab`

(C)

`b/a`

(D)

`(a-b)/(1+ab)`

Solution:

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

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

and `2 tan^(-1)x=tan^(-1)x((2x)/(1-x^2))]`

`:. 2 tan^(-1) a-2 tan^(-1) b=2 tan^(-1) x`

`=> tan^(-1)a-tan^(-1) b=tan^(-1) x`

`=> tan^(-1) ((a-b)/(1+ab))=tan^(-1)x`

`[ ∵ tan^(-1) x-tan^(-1) y=tan^(-1) ((x-y)/(1+xy))]`

`:. x=(a-b)/(1+ab)`
Correct Answer is `=>` (D) `(a-b)/(1+ab)`
Q 2446201173

The value of `2 tan^(- 1) (cosec tan^(- 1) x - tan cot^(- 1) x)`
is
UPSEE 2014
(A)

`tan^(-1) x`

(B)

`tan x`

(C)

`cot x`

(D)

`cosec^(- 1)x`

Solution:

`2 tan^(- 1) (cosec tan^(- 1) x- tan cot^(- 1) x)`

`= 2 tan^(- 1) [ (cosec { cosec^(-1) sqrt(1-x^2)/x})-(tan {tan^(- 1) (1/x)})]`

` = 2 tan^(-1) [sqrt(1-x^2)/x - 1/x] = 2 tan^(-1) [(sqrt(1-x^2) - 1)/x]`

`= 2 tan^(- 1) [ (sec theta - 1)/(tan theta) ]` (put `x = tan theta` )

`=2 tan^(-1) [(1 - cos theta )/(sin theta)]`

`= 2 tan^(-1) [( 2 sin^2 \ theta/2)/(2 sin \ theta/2 . cos \ theta/2)]`

`= 2 tan^(-1) (tan \ theta/2) = 2 · theta/2 = theta = tan^(-1) x`
Correct Answer is `=>` (A) `tan^(-1) x`
Q 2460734615

If `tan^(-1) 2` and `tan^(-1) 3` are two angles of a triangle,
then what is the third angle?
NDA Paper 1 2012
(A)

`tan^(-1) 2`

(B)

`tan^(-1) 4`

(C)

`pi/4`

(D)

`pi/3`

Solution:

Given that, `tan^(-1)2` and `tan^(-1) 3` are two angles of a
triangle. Let `alpha` be the third angle of the triangle, then -
In `Delta ABC, tan^(-1) 2 + tan^(-1) 3 + alpha= 180°`

`=> tan^(-1) ((2+3)/(1-2 *3))+alpha=180^o`

`[ ∵ tan^(-1) x+tan^(-1)y=tan^(-1) ((x+y)/(1-xy))]`

`=>tan^(-1) (5/(1-6))+alpha=180^o=> tan^(-1) (5/(-5))+alpha=180^o`

`=> tan^(-1) (-1)+alpha=180^o=>alpha=180^o-tan^(-1)(-1)=pi-(3 pi)/4`

`:. alpha=pi/4`
Correct Answer is `=>` (C) `pi/4`
Q 2270191916

The value of `tan ((2 tan^(-1)) 1/5 - pi /4)` is
NDA Paper 1 2015
(A)

` - 7/(17)`

(B)

` 5/(16)`

(C)

`5/4`

(D)

` 7/(17)`

Solution:

`tan ((2 tan^(-1)) 1/5 - pi /4)` ..........(1)

` ∵(2 tan^(-1)) 1/5 = tan^(-1) ( (2xx1/5)/(1- (1/5)^2) )`

`[∵ 2 tan^(-1) x = tan^(-1) (2x)/(1-x^2)`, for ` |x| <1]`

` =tan^(-1)( 1/5)`

Now, from Eq. (i),

` tan ((tan^(-1)) 5/(12) - pi /4) = ( tan (tan^(-1) 5/(12)) - tan pi/4)/ ( 1 + tan (tan^(-1) 5/(12)) . tan pi/4)`

` [∵ tan (A-B) = (tan A - tan B)/(1 + tan A . tan B)]`

` = ( 5/(12) - 1)/(1+ 5/(12) (1)) = (-7)/(17)`
Correct Answer is `=>` (A) ` - 7/(17)`
Q 1618180000

Consider `x = 4 (tan^(-1))(1/5)`

`y = (tan^(-1)) (1/(70) )` and `z = (tan^(-1)) (1/(99) )`

What is `x- y + z` equal to?
NDA Paper 1 2015
(A)

` pi/2 `

(B)

`pi/3`

(C)

`pi/6`

(D)

`pi/4`

Solution:

Given, ` x = 4 (tan^(-1)) (1/5) = 2 . 2 (tan^(-1)) (1/5)`

` =( 2 . (tan^(-1))) ( 2 // 5)/( 24//25) = 2 (tan^(-1)) 5//12`

` = (2 (tan^(-1))) (10//12)/(1 - (25)/(144)) = ( (tan^(-1)) )(10//12)/( 119 // 144)`

` = (2 (tan^(-1))) ( 2 // 5)/( 119//144) = (tan^(-1)) ( 144 xx 10)/( 12 xx 119)`

` = x = (tan^(-1)) ((120)/(119))`

Now, `x - y = (tan^(-1)) (120)/(119) - (tan^(-1)) 1/(70) = ( (tan^(-1))) ((120)/(119) - 1/(70))/( 1+ (120)/(119) xx 1/(70) )`

` = (tan^(-1)) (120 xx 70 - 119)/( 119 xx 70 + 120)`

` = (tan^(-1)) ( 8400 - 119) /(8330 - 120) = (tan^(-1)) ( 8281)/(8450)`

` :. x - y + z = (tan^(-1)) ( 8281)/(8450) + (tan^(-1)) 1/(99)`

` = (tan^(-1)) (( 8281)/(8450) + 1/(99) )/ ( 1 - (8281 xx 1)/(8450 xx 99) )`

` = (tan^(-1)) ( 99 xx 8281 + 8450 )/( 8450 xx 99 - 8281)`

` = (tan^(-1)) ( 828269)/(828269) = (tan^(-1)) 1`

` = (tan^(-1)) ( tan pi/4 ) = pi/4`
Correct Answer is `=>` (D) `pi/4`

Problems based on Property 5

(i) `2tan^-1 x = sin^-1 \ \( 2x) /( 1 + x^2) , | x| <=1`

(ii) `2tan^-1 x = cos^-1 \ \( 1- x^2 ) /( 1 +x^2 ) , x >= 0`

(iii) `2 tan^-1 x = tan^-1 \ \ (2 x) /(1 -x^2 ) . -1 < x < 1`

Let `tan^-1 x = y`, then `x = tan y`.

Now `sin^-1 \ \ ( 2 x) /( 1 + x^2) = sin^-1 \ \ ( 2 tan y) /( 1 + tan^2 y)`

`= sin^-1 (sin 2y) = 2y = 2tan^-1 x`

Also `cos^-1 \ \ ( 1 - x^2 ) /( 1 +x^2) = cos^-1 \ \ ( 1 - tan^2y)/( 1 + tan^2y) = cos ^-1 \ \ ( cos 2y ) = 2y = 2 tan ^-1 x`

(iii) Can be worked out similarly.

We now consider some examples.
Q 2420745611

What is the value of `sin^(-1) (4/5)+2 tan^(-1) (1/3)` ?
NDA Paper 1 2010
(A)

`pi/3`

(B)

`pi/2`

(C)

`pi/4`

(D)

`pi/6`

Solution:

`sin^(-1) (4/5)+2tan^(-1)(1/3)`

`=tan^(-1)((4//5)/(sqrt(1-16//25))+tan^(-1)( (2//3)/(1-1//9))`

`[ ∵ sin^(-1) x=tan^(-1) (x/(sqrt(1-x^2)))` and `2tan^(-1) x=tan^(-1)((2x)/(1-x^2))]`

`=tan^(-1)((4//5)/(3//5))+tan^(-1)((2//3)/(8//9))`

`=tan^(-1)(4/3)+tan^(-1)(3/4) ` `[ ∵ tan^(-1) x=cot^(-1) (1/x)]`

`tan^(-1)(4/3)+cot^(-1)(4/3)=pi/2` `( ∵ tan^(-1)+cot^(-1)x=pi/2)`
Correct Answer is `=>` (B) `pi/2`
Q 2480134017

If `sin^(-1) ((2a)/(1+a^2))+sin^(-1) ((2b)/(1+b^2))=2 tan^(-1) x` then `x` is equal to
NDA Paper 1 2013
(A)

`(a-b)/(1+ab)`

(B)

`(a-b)/(1-ab)`

(C)

`(2ab)/(a+b)`

(D)

`(a+b)/(1-ab)`

Solution:

Given that,

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

`=> 2 tan^(-1) a+2 tan^(-1) b=2 tan^(-1) x`

`=> (tan^(-1)a+tan^(-1) b)=tan^(-1) x`

`=>tan^(-1) ((a+b)/(1-ab))=tan^(-1) x`

`[ ∵ tan^(-1) x+tan^(-1)y=tan^(-1) ((x+y)/(1-xy))]`

`:. x=(a+b)/(1-ab)` where `a > 0, b > 0 .`
Correct Answer is `=>` (D) `(a+b)/(1-ab)`
Q 2480123917

What is `tan^(-1) (1/2) +tan^(-1) (1/3)` equal to?
NDA Paper 1 2013
(A)

`pi/2`

(B)

`pi/3`

(C)

`pi/4`

(D)

`pi/6`

Solution:

`tan^(-1) (1/2) +tan^(-1) (1/3)= tan^(-1)((1/2+1/3)/(1-1/6))`

`[ ∵ tan^(-1) x+tan^(-1)y=tan^(-1) ((x+y)/(1-xy))]`

`=tan^(-1) (5/6 xx 6/5)=tan^(-1)(1)=tan^(-1)tan (pi/4)=pi/4`
Correct Answer is `=>` (C) `pi/4`
Q 2201301228

Consider the following
1. ` sin ^(-1) (4/5) + sin^(-1) (3/5) = pi/2`

2. `tan ^(-1) sqrt(3) +tan^( -1) 1 = -tan ^(- 1) (2 + sqrt(3))`

Which of the above is/are correct?

NDA Paper 1 2015
(A)

Only `1`

(B)

Only `2`

(C)

Both `1` and `2`

(D)

Neither `1` nor `2`

Solution:

1. ` LHS = sin ^(-1) (4/5) + sin^ (-1) (3/5)`

` = sin ^(-1) (4/5) + cos^ (-1) (3/5)` `\ \ \ \ \ \ \ \[ ∵ sin ^(-1) (3/5) + cos^ (-1) (4/5)]`

` = pi/2 = RHS`

2. `LHS = tan ^(-1) sqrt(3) + tan ^(- 1) 1`

` = tan ^(-1) ( (sqrt(3) +1)/(1- sqrt(3)) xx ( 1+ sqrt(3))/( 1+ sqrt(3) ))`

` = tan ^(-1) ( ( 3+1 + 2sqrt(3))/(1-3)) = tan ^(-1) ((2+sqrt(3))/(-1))`

` = - tan ^(-1) ( 2 + sqrt(3))`

Hence, both the statements are correct.
Correct Answer is `=>` (C) Both `1` and `2`

Problems based on Property 6

`sin^(-1)x + sin^(-1)y = sin^(-1)(xsqrt(1-y^2) + y sqrt(1-x^2))`

`sin^(-1)x - sin^(-1)y = sin^(-1)(xsqrt(1-y^2) - y sqrt(1-x^2))`

`cos^(-1)x + cos^(-1)y = cos^(-1)(xy + sqrt(1-x^2) .sqrt(1-x^2))`

`cos^(-1)x + cos^(-1)y = cos^(-1)(xy - sqrt(1-x^2) .sqrt(1-x^2))`
Q 2480034817

What is the value of `cos {cos^(-1) \ 4/5+cos^(-1) \ 12/13}` ?
NDA Paper 1 2012
(A)

`63/65`

(B)

`33/65`

(C)

`22/65`

(D)

`11/65`

Solution:

`cos {cos^(-1) \ 4/5+cos^(-1) \ 12/13}`

`=cos cos ^(-1) { 4/5 * 12/13- sqrt(1-(4/5)^2) * sqrt(1-(12/13)^2)}`

`[ ∵ cos^(-1) x +cos^(-1)y=cos^(-1){xy-sqrt(1-x^2)*sqrt(1-y^2)]`

`=48/65-sqrt(1-16/25)*sqrt(1-144/169)`

`=48/65-sqrt(9/25) * sqrt(25/169)`

`=48/65-3/5 * 5/13=48/65-3/13=(48-15)/65=33/65`
Correct Answer is `=>` (B) `33/65`
 
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