Here we are providing Class 12 Maths Important Extra Questions and Answers Chapter 2 Inverse Trigonometric Functions. Class 12 Maths Important Questions are the best resource for students which helps in Class 12 board exams.

## Class 12 Maths Chapter 2 Important Extra Questions Inverse Trigonometric Functions

### Inverse Trigonometric Functions Important Extra Questions Very Short Answer Type

Question 1.

Find the principal value of sin^{-1} ( \(\frac { 1 }{ 2 }\) )

Solution:

\(\sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)

Hence, the principal value of sin^{-1} ( \(\frac { 1 }{ 2 }\) ) is \(\frac{\pi}{6}\)

Question 2.

What is the principal value of:

cos^{-1} (cos \(\frac{2 \pi}{3}\) + sin^{-1} (sin \(\frac{2 \pi}{3}\) ) ?

Solution:

Question 3.

Find the principal value of:

tan^{-1} (√3)- sec^{-1} (-2). (A.I.C.B.S.E. 2012)

Solution:

Question 4.

Evaluate : tan ^{-1} ( 2 cos (2 sin^{-1} ( \(\frac{1}{2}\) )))

Solution:

Question 5.

Find the value of tan^{-1}(√3) – cot^{-1}(-√3). (C.B.S.E. 2018)

Solution:

tan^{-1}(√3) – cot^{-1}(—√3)

\(=\frac{\pi}{3}-\left(\pi-\frac{\pi}{6}\right)=-\frac{\pi}{2}\)

Question 6.

If sin^{-1} ( \(\frac { 1 }{ 3 }\) ) + cos^{-1} x = \(\frac{\pi}{2}\), then find x.(C.B.S.E. 2010C)

Solution:

sin^{-1} ( \(\frac { 1 }{ 3 }\) ) + cos^{-1} x = \(\frac{\pi}{2}\)

⇒ x = 1/3

[sin^{-1} x + cos^{-1} x = \(\frac{\pi}{2}\)

Question 7.

If sec^{-1} (2) + cosec^{-1} (y) = \(\frac{\pi}{2}\) , then find y.

Solution:

sec^{-1} (2) + cosec^{-1} (y) = \(\frac{\pi}{2}\)

⇒ y = 2 [∵ sec^{-1} x + cosec^{-1} x= \(\frac{\pi}{2}\) ]

Question 8.

Write the value of sin [ \(\frac{\pi}{3}\) – sin ^{-1} ( \(\frac { -1 }{ 2 }\) ) ]

Solution:

sin [ \(\frac{\pi}{3}\) – sin^{-1} ( \(\frac { -1 }{ 2 }\) ) ]

= sin [ \(\frac{\pi}{3}\) + sin^{-1} ( \(\frac { 1 }{ 2 }\) ) ]

[∵ sin^{-1} (-x) = -sin^{-1}x]

= sin ( \(\frac{\pi}{3}+\frac{\pi}{6}\) ) = sin \(\frac{\pi}{2}\) = 1

Question 9.

Prove the following:

Solution:

Question 10.

If tan^{-1} x + tan^{-1}y = \(\frac{\pi}{4}\) , xy < 1, then write the value of the x + y + xy (A.I.C.B.S.E. 2014)

Solution:

We have tan^{-1} x + tan^{-1}y = \(\frac{\pi}{4}\)

Hence x + y + xy = 1.

Question 11.

Prove that: 3 sin^{-1} x = sin^{-1}(3x – 4x^{2});

x ∈ [\(\frac { -1 }{ 2 }\) , \(\frac { 1 }{ 2 }\)] (C.B.S.E 2018)

Solution:

To prove: 3 sin^{-1} x = sin^{-1}(3x – 4x^{2})

Put sin^{-1} x = θ

so that x = sin θ.

RHS = sin^{-1} (3 sin θ – 4 sin^{3} θ)

= sin^{-1} (sin 3θ) = 3θ = 3 sin^{-1}x = LHS.

### Inverse Trigonometric Functions Important Extra Questions Short Answer Type

Question 1.

Express sin^{-1} ( \(\frac{\sin x+\cos x}{\sqrt{2}}\) )

Where \(-\frac{\pi}{4}\) < x < \(\frac{\pi}{4}\), in the simples form.

Solution:

Question 2.

Prove that :

\(\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{56}{65}\) (A.I.C.B.S.E. 2019; C.B.S.E. 2010)

Solution:

Question 3.

Prove that :

sin ^{-1} \(\frac{8}{17}\) + cos ^{-1} \(\frac{4}{5}\) = cos^{-1} \(\frac{36}{77}\) (A.I.C.B.S.E. 2019)

Solution:

L.H.S = sin^{-1} \(\frac{8}{17}\) + cos^{-1} \(\frac{4}{5}\)

= tan^{-1} \(\frac{8}{15}\) + tan^{-1}\(\frac{3}{4}\)

= R.H.S

Question 4.

Solve the following equation:

\(\tan ^{-1}\left(\frac{x+1}{x-1}\right)+\tan ^{-1}\left(\frac{x-1}{x}\right)=\tan ^{-1}(-7)\) (A.I.C.B.S.E. 2019; C)

Solution:

⇒ 2x^{2} – 8x + 8 = 0

⇒ x^{2} – 4x + 4 = 0

⇒ (x – 2)^{2} = 0.

Hence, x = 2.

Question 5.

Solve the following equation:

2 tan^{-1}(sin x) = tan^{-1} (2 sec x), x ≠ \(\frac{\pi}{2}\) (C.B.S.E. (F) 2012)

Solution:

2 tan^{-1}(sinx) = tan^{-1} (2 secx)

⇒ tan ^{-1} (2sec x tan x) = tan^{-1} (2sec x)

⇒ 2 sec x tan x = 2 sec x

tan x = 1 [∵ sec x ≠ 0 ]

Hence, x= \(\frac{\pi}{2}\)

Question 6.

Solve the following equation

cos (tan^{-1} x) = sin ( cot ^{-1}\(\frac { 3 }{ 4 }\) )

(A.I.C.B.S.E. 2013)

Solution:

We have:

cos (tan^{-1} x) = sin ( cot ^{-1}\(\frac { 3 }{ 4 }\) )

cos (tan^{-1} x) = sin ( sin ^{-1}\(\frac { 4 }{ 5 }\) )

cos (tan^{-1} x) = \(\frac { 4 }{ 5 }\)

tan^{-1} x = cos^{-1}\(\frac { 4 }{ 5 }\)

⇒ tan^{-1} x = tan ^{-1}\(\frac { 3 }{ 4 }\)

Hence x = \(\frac { 3 }{ 4 }\)

Question 7.

Prove that

3cos^{-1} x = cos^{-1} (4x^{3} – 3x), x ∈ [ \(\frac { 1 }{ 2 }\) , 1 ]

Solution:

Put x = cos θ in RHS

As 1/2 ≤ x ≤ 1

RHS = cos^{-1} (4cos^{3} θ – 3cos θ),

= cos^{-1} (cos 3θ) = 3θ = 3cos^{-1} x = L.H.S

### Inverse Trigonometric Functions Important Extra Questions Long Answer Type 1

Question 1.

prove that \(\frac { 1 }{ 2 }\) ≤ x ≤ 1, then

\(\cos ^{-1} x+\cos ^{-1}\left[\frac{x}{2}+\frac{\sqrt{3}-3 x^{2}}{2}\right]=\frac{\pi}{3}\) (CBSE. Sample paper 2017 – 18)

Solution:

Question 2.

Find the value of :

\(\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right)\)

Solution:

Question 3.

Prove that :

tan ^{-1}( \(\frac{1}{2}\) ) + tan ^{-1} ( \(\frac{1}{5}\) ) + tan ^{-1}( \(\frac{1}{8}\) ) = \(\frac{\pi}{4}\) (C.B.S.E. 2013: A.I.C.B.S.E. 2011)

Solution:

Question 4.

Solution:

Question 5.

Solution:

Put cos ^{-1} (a/b) = θ so that cos θ = a/b.

Question 6.

Solution:

Question 7.

( A.I.C.B.S.E. 2016)

Solution:

Question 8.

\(\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}\) (C.B.S.E. 2016)

Solution:

Question 9.

Solve for x : 2 tan^{-1} (cos x ) = tan^{-1}(2 cosec x). (C.B.S.E. 2016)

Solution:

2 tan^{-1} (cos x ) = tan^{-1} (cos x) + tan^{-1} (cos x)

= tan^{-1} (2cot x cosec x) …(1)

Now 2 tan^{-1}(cos x) = tan^{-1}(2 cosec x)

⇒ tan^{-1}(2 cot x cosec x) = tan^{-1} (2cosec x)

[Using (1)]

⇒ 2 cot x cosec x = 2 cosec x

⇒ cotx cosecx = cosec x

⇒ sin x = tan x sin x

⇒either sin x = 0 or tan x = 1.

Hence, x = nπ ∀ n ∈ Z or x

= πm + \(\frac{\pi}{4}\) ∀ m ∈ Z

Question 10.

If tan^{-1} \(\left(\frac{x-2}{x-4}\right)\) + tan^{-1} \(\left(\frac{x+2}{x+4}\right)\) = \(\frac{\pi}{4}\) find the value of ‘x’ (A.I.C.B.S.E. 2014)

Solution:

We have

⇒ 2x^{2} – 16 = -12

⇒ 2x^{2} = 16 – 12

⇒ 2x^{2} = 4

⇒ x^{2} = 2

Hence x = ±√2

Question 11.

If tan^{-1} \(\left(\frac{x-3}{x-4}\right)\) + tan^{-1} \(\left(\frac{x+3}{x+4}\right)\) = \(\frac{\pi}{4}\) find the value of ‘x’ (A.I.C.B.S.E. 2017)

Solution:

⇒ 2x^{2} – 24 = -7

⇒ 2x^{2} = 17

⇒ 2x^{2} = 4

⇒ x^{2} = 17/2

Hence x = \(\pm \sqrt{\frac{17}{2}}\)

Question 12.

Prove that : \(\tan ^{-1} \sqrt{x}=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right)\)

x∈ [0,1] (C.B.S.E. 2019C)

Solution:

Let tan ^{-1} √x = θ

So that √x = tan θ i.e. x = tan^{2}θ

Question 13.

Solve for x : tan ^{-1} (2x) + tan ^{-1} (3x) = \(\frac{\pi}{4}\) (C.B.S.E. 2019)

Solution:

The given equation is

tan ^{-1} (2x) + tan ^{-1} (3x) = \(\frac{\pi}{4}\) …. (1)

⇒ 5x = 1 – 6x^{2}

⇒ 6x^{2} + 5x – 1 = 0

⇒ x = -1 or x = \(\frac{1}{6}\)

Hence, x = \(\frac{1}{6}\)

[∵ x = -1 does not satisfy (1)]

Question 14.

Solve : tan^{-1} 4x + tan^{-1} 6x = \(\frac{\pi}{4}\) (C.B.S.E. 2019)

Solution:

We have tan^{-1} 4x + tan^{-1} 6x = \(\frac{\pi}{4}\)

⇒ 10x = 1 – 24x^{2}

⇒ 24x^{2} + 10x – 1 = 0

⇒ 24x^{2}+ 12x – 2x – 1 = 0

⇒ 12x(2x + 1) – 1 (2x + 1) = 0

⇒ (2x+1)(12x – 1) = 0

⇒ 2x + 1 = 0 or 12x – 1 = 0

x = \(-\frac{1}{2}\) or x = \(\frac{1}{2}\).

Hence, x = \(-\frac{1}{2}\) or \(\frac{1}{12}\)

As x = \(-\frac{1}{2}\) does not satisfy the given equation.

Hence x = \(-\frac{1}{12}\)

Question 15.

If (tan^{-1} x )^{2} + (cot^{-1} x )^{2} = \(\frac{5 \pi^{2}}{8}\) , then find ‘x’ (C.B.S.E. 2015)

Solution:

We have

(tan^{-1} x )^{2} + (cot^{-1} x )^{2} = \(\frac{5 \pi^{2}}{8}\)

Put tan^{-1} x = t so that cot^{-1}x = \(\frac{\pi}{2}\) – t

∴ (1) become : t^{2} + (\(\frac{\pi}{2}\) – t)^{2} = \(\frac{5 \pi^{2}}{8}\)

When tan^{-1}x = \(\frac{3 \pi}{4}\)

then x = tan \(\frac{3 \pi}{4}\) = -1.

When tan^{-1} x = \(-\frac{\pi}{4}\),

then x = tan(\(-\frac{\pi}{4}\)) = -1.

Hence, x = -1.

Question 16.

Write : tan ^{-1}\(\frac{1}{\sqrt{x^{2}-1}}\) , |x| > 1 in the simplest form.

Solution:

Put x = sec θ

Question 17.

Prove that

(C.B.S.E. 2012)

Solution:

Question 18.

Prove that :

tan^{-1}x + tan^{-1} \(\frac{2 x}{1-x^{2}}\) = tan ^{-1}\(\frac{3 x-x^{3}}{1-3 x^{2}}\) (N.C.E.R.T

Solution:

Put x = tan θ so that θ = tan ^{-1} x

= 3θ = 3 tan^{-1}x = tan^{-1}x + 2tan^{-1}x

= tan^{-1}x + tan^{-1}\(\frac{2 x}{1-x^{2}}\)

= L.H.S.

Question 19.

Simplify : tan^{-1} [ \(\left[\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right]\)]

If \(\frac{a}{b}\) tan x > -1 (N.C.E.R.T)

Solution:

Question 20.

Prove that tan ^{-1} [ \(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\) ]

= \(\frac{\pi}{4}+\frac{1}{2}\) cos^{-1} x ; \(-\frac{1}{\sqrt{2}}\) ≤ x ≤ 1. (C.B.S.E. 2019C)

Solution:

Question 21.

Show that

(N.C.E.R.T.A.I. CBSE 2014)

Solution:

Question 22.

cos[tan^{-1}{sin (cot^{-1}x)}] = \(\sqrt{\frac{1+x^{2}}{2+x^{2}}}\)

(A.I. C.B.S.E. 2010)

Solution:

Put cot^{-1} = θ so that x = cot θ

∴ sin θ = \(\frac{1}{\sqrt{1+x^{2}}}\)

Question 23.

Find the value of sin (2 tan^{-1} 1/4) + cos (tan^{-1 }2√2) (C.B.S.E. Sample Paper 2018 – 2019)

Solution:

Put tan^{-1} 1/4 = θ so that θ = 1/4

Now, sin 2θ = \(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\)

To evaluate cos(tan^{-1 }2√2) :

Put tan^{-1}2√2 = Φ

So that tan Φ = 2√2

cos Φ = 1/3

Hence, sin (2tan^{-1}(1/4)) + cos (tan^{-1}2√2)

= sin 2θ + cos Φ = \(\frac{8}{17}+\frac{1}{3}=\frac{41}{51}\)

[Using (1) and (2)]

Question 24.

Solve for x:

tan^{-1}(x – 1) + tan^{-1}x + tan^{-1}(x + 1) = tan^{-1}(3x). (A.I.C.B.S.E. 2016)

The given equation is:

tan^{-1}(x —1) + tan^{-1}(x) + tan^{-1}(x + 1) = tan^{-1}3x

tan^{-1}(x— 1)+tan^{-1}(x+ 1) = tan^{-1}3x – tan^{-1}

⇒ 1 + 3x^{2} = 2 – x^{2}

⇒ 4x^{2} = 1

⇒ x^{2} = 1/4

⇒ x = \(\pm \frac{1}{2}\)

Hence, x = 0, \(\pm \frac{1}{2}\)