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CBSE Important Questions Class 12 Maths Chapter 2 – Inverse Trigonometric Functions

Class 12 Maths Chapter 2 Inverse Trigonometric Functions is part of the CBSE Class 12 Maths syllabus. It explains how the trigonometric functions can be inverted once their domains are restricted. Students learn the domain and range (principal value branch) of each inverse function, how to find principal values, and the standard properties and identities used to simplify expressions.

These CBSE important questions for Class 12 Maths Chapter 2 cover the domain of a combined inverse trigonometric expression, finding principal values, simplifying sin⁻¹(sin x) and tan⁻¹(tan x) when the angle is outside the principal branch, case studies on angles of elevation at a temple and a billboard, and assertion-reason questions. Every question has a clear step-by-step solution in simple English with figures, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. The domain of the function f(x) = sin⁻¹x + cos⁻¹x + cosec⁻¹x + cot⁻¹x + tan⁻¹x is
(a) R − [−1, 1]
(b) {−1, 1}
(c) [−1, 1]
(d) R − (−1, 1)

Answer: (b) {−1, 1}
Take the domain of each piece:
sin⁻¹x and cos⁻¹x need x ∈ [−1, 1]
cosec⁻¹x needs |x| ≥ 1, that is x ∈ R − (−1, 1)
tan⁻¹x and cot⁻¹x are defined for all real x
The domain of f is the intersection of all these.
[−1, 1] ∩ (R − (−1, 1)) = {−1, 1}

Q2. Write the principal value branch (domain and range) of each inverse trigonometric function.

Function	Domain	Range (principal branch)
sin⁻¹x	[−1, 1]	[−π/2, π/2]
cos⁻¹x	[−1, 1]	[0, π]
tan⁻¹x	R	(−π/2, π/2)
cot⁻¹x	R	(0, π)
sec⁻¹x	R − (−1, 1)	[0, π] − {π/2}
cosec⁻¹x	R − (−1, 1)	[−π/2, π/2] − {0}

Q3. Find the value of sin⁻¹(sin 2π/3) + cos⁻¹(cos 4π/3).

Answer: Neither angle lies in its principal branch, so we reduce each one first.
For sin⁻¹(sin 2π/3): Write 2π/3 = π − π/3, and sin(π − θ) = sin θ.
= sin⁻¹(sin π/3) = π/3 (since π/3 ∈ [−π/2, π/2])
For cos⁻¹(cos 4π/3): Write 4π/3 = 2π − 2π/3, and cos(2π − θ) = cos θ.
= cos⁻¹(cos 2π/3) = 2π/3 (since 2π/3 ∈ [0, π])
Sum = π/3 + 2π/3 = π

Q4. Find the principal value of sin⁻¹(−√3/2).

Answer: We know sin(π/3) = √3/2, so sin(−π/3) = −√3/2.
sin⁻¹(−√3/2) = −π/3
Check that −π/3 lies in the principal branch [−π/2, π/2] — it does.
So the principal value is −π/3.

Q5. Find the value of sin⁻¹(sin 4π/5) and tan⁻¹(tan 5π/6).

Answer: sin⁻¹(sin 4π/5): Since 4π/5 is not in [−π/2, π/2], write 4π/5 = π − π/5.
Using sin(π − θ) = sin θ:
= sin⁻¹(sin π/5) = π/5
tan⁻¹(tan 5π/6): Since 5π/6 is not in (−π/2, π/2), write 5π/6 = π − π/6.
Using tan(π − θ) = −tan θ:
= tan⁻¹(−tan π/6) = −π/6

Q6. Two individuals stand on opposite sides of a temple that is 60√3 m high. They observe the top of the temple at angles of elevation θ and φ respectively. The distance between the two is 240 m and the distance from the first person to the temple is 60 m. Find θ and φ.

Two people standing on opposite sides of a temple observing its top at angles theta and phi

Answer: Finding θ: In the right triangle formed by the first person P:
tanθ = height / OP = 60√3 / 60 = √3
Since tan 60° = √3:
θ = 60°, that is θ = tan⁻¹(√3) = π/3
Finding φ: The distance of the second person from the temple is
OR = PR − OP = 240 − 60 = 180 m
tanφ = 60√3 / 180 = √3/3 = 1/√3
Since tan 30° = 1/√3:
φ = 30°, that is φ = tan⁻¹(1/√3) = π/6

Q7. A person P stands 30 m from the base of a tower. A billboard is to be placed at height X = 15 m. The angle of elevation to point Y is twice that to X, and to point Z is three times that to X. Find ∠XPQ, ∠YPQ and ∠ZPQ.

A tower with three marked heights X, Y and Z observed from a point P on the ground

Answer: ∠XPQ: Let this angle be θ.
tanθ = XQ / PQ = 15/30 = 1/2
∠XPQ = tan⁻¹(1/2)
∠YPQ = 2θ: Using tan 2θ = 2tanθ / (1 − tan²θ)
= 2(1/2) / (1 − 1/4)
= 1 / (3/4) = 4/3
∠YPQ = tan⁻¹(4/3)
∠ZPQ = 3θ: Using tan 3θ = (3tanθ − tan³θ) / (1 − 3tan²θ)
= [3(1/2) − 1/8] / [1 − 3(1/4)]
= (11/8) / (1/4) = 11/2
∠ZPQ = tan⁻¹(11/2)

Q8. Read the statements and select the correct option.
Assertion (A): cos⁻¹(−x) = π − cos⁻¹x
Reason (R): The cosine function is odd.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true, but R is false
(d) A is false, but R is true

Answer: (c) A is true, but R is false
The Assertion is a standard and correct property of inverse trigonometric functions.
The Reason is wrong: the cosine function is an even function, since cos(−x) = cos x. It is sine and tangent that are odd.

Q9. Read the statements and select the correct option.
Assertion (A): sin⁻¹(sin 2) = 2
Reason (R): sin⁻¹(sin x) = x for all real values of x.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true, but R is false
(d) A is false, but R is true

Answer: Both the Assertion and the Reason are false.
sin⁻¹(sin x) = x holds only when x lies in [−π/2, π/2], which is about [−1.57, 1.57].
Here x = 2, which is outside this range. The correct value is sin⁻¹(sin 2) = π − 2.

Q10. Write the main properties of inverse trigonometric functions.

Answer: • sin⁻¹(−x) = −sin⁻¹x and tan⁻¹(−x) = −tan⁻¹x
• cos⁻¹(−x) = π − cos⁻¹x and cot⁻¹(−x) = π − cot⁻¹x
• sin⁻¹x + cos⁻¹x = π/2
• tan⁻¹x + cot⁻¹x = π/2
• sec⁻¹x + cosec⁻¹x = π/2
• sin⁻¹(1/x) = cosec⁻¹x, and similarly for the other reciprocal pairs

Q11. Find the value of tan⁻¹(1) + cos⁻¹(−1/2) + sin⁻¹(−1/2).

Answer: Work out each term separately.
tan⁻¹(1) = π/4
cos⁻¹(−1/2) = π − cos⁻¹(1/2) = π − π/3 = 2π/3
sin⁻¹(−1/2) = −sin⁻¹(1/2) = −π/6
Adding all three:
= π/4 + 2π/3 − π/6
Taking the LCM 12: = (3π + 8π − 2π)/12
= 9π/12 = 3π/4

Q12. Prove that 2 tan⁻¹(1/2) + tan⁻¹(1/7) = tan⁻¹(31/17).

Answer: First use 2 tan⁻¹x = tan⁻¹[2x / (1 − x²)] with x = 1/2:
2 tan⁻¹(1/2) = tan⁻¹[ (2 × 1/2) / (1 − 1/4) ]
= tan⁻¹[ 1 / (3/4) ] = tan⁻¹(4/3)
Now use tan⁻¹A + tan⁻¹B = tan⁻¹[(A + B)/(1 − AB)] with A = 4/3 and B = 1/7:
= tan⁻¹[ (4/3 + 1/7) / (1 − 4/21) ]
= tan⁻¹[ (28 + 3)/21 ÷ (17/21) ]
= tan⁻¹[ 31/21 × 21/17 ]
= tan⁻¹(31/17). Hence proved.

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Q1.

$Evaluate \cot (\tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{1}{8})$

opt.

$\begin{array}{l} We have \\ \cot [\tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{1}{8}] \\ = \cot [\tan^{- 1} 1] \end{array}$

ans.

$\begin{array}{l} We have \\ \cot [\tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{1}{8}] \\ = \cot [\tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{1}{8}] \\ = „ \cot „ [\tan^{1} (\frac{\frac{1}{5} + \frac{1}{7}}{1 \frac{1}{5} Ã— \frac{1}{7}}) + \tan^{1} (\frac{\frac{1}{3} + \frac{1}{8}}{1 \frac{1}{3} Ã— \frac{1}{8}})] \\ = „ \cot „ [\tan^{1} (\frac{\frac{7 + 5}{35}}{1 \frac{1}{35}}) + \tan^{1} (\frac{\frac{8 + 3}{24}}{1 \frac{1}{24}})] \\ = „ \cot „ [\tan^{1} (\frac{\frac{12}{35}}{\frac{34}{35}}) + \tan^{1} (\frac{\frac{11}{24}}{\frac{23}{24}})] \\ = „ \cot „ [\tan^{1} (\frac{12}{34}) + \tan^{1} (\frac{11}{23})] \\ = „ \cot „ [\tan^{1} (\frac{6}{17}) + \tan^{1} (\frac{11}{23})] \\ = „ \cot „ + [\tan^{1} (\frac{\frac{138 + 187}{17 Ã— 23}}{\frac{391 66}{17 Ã— 23}})] \\ = „ \cot „ + [\tan^{1} (\frac{325}{325})] \\ = „ \cot „ + [\tan^{1} (1)] \\ = „ \cot „ \frac{Ï}{4} \\ = „ 1 \end{array}$

Q2-

$Solve for x \sin^{- 1} (1 - x) - 2 \sin^{- 1} x = \frac{}{2} .$

opt-

$\begin{array}{l} We have \\ \sin^{1} (1 x) 2 \sin^{1} x = \frac{}{2} \\ \sin^{1} (1 x) = \frac{}{2} + 2 \sin \end{array}$

ans-

$\begin{array}{l} We have \\ \sin^{1} (1 x) 2 \sin^{1} x „ = \frac{Ï}{2} \\ „ \sin^{1} „ (1 x) „ = „ \frac{Ï}{2} „ + „ 2 „ \sin^{1} x \\ „ (1 x) „ = „ \sin „ (\frac{Ï}{2} + 2 \sin^{1} x) \\ „ (1 x) „ = „ \cos „ (2 „ \sin^{1} x) \\ „ \cos^{1} (1 x) „ = „ 2 \sin^{1} x \\ „ \sin^{1 „} \sqrt{1 {(1 x)}^{2}} = „ \sin^{1} (2 x \sqrt{1 x^{2}}) \\ \sqrt{1 {(1 x)}^{2}} „ = „ (2 x „ \sqrt{1 x^{2}}) \\ „ „„ On squaring both sides \\ 1 - {(1 x)}^{2} = 4 x^{2} (1 x^{2}) \\ „ 1 1 x^{2} + 2 x = „ 4 x^{2} 4 x^{4} \\ „ 4 x^{4} 5 x^{2} + 2 x „ = „ 0 \\ „ x „ (4 x^{3} 5 x + 2) „ = „ 0 \\ ’ „ x = 0 or \\ 4 x^{3} 5 x + 2 = „ 0 \\ Put x = . .. -1 0 1 2 . \dots we get no real vlue of x \\ satisfy above cubic equation in [1 „ 1] \\ Hence x = 0 is only solution . \end{array}$

Q3-

$Find the value of \cos^{1} (\cos \frac{13}{6}) .$

opt-

$\begin{array}{l} \cos^{1} (\cos \frac{13}{6}) = \frac{13}{6} \\ But \frac{13}{6} [0] \\ \cos (\frac{13}{6}) = \cos (2) \end{array}$

ans-

$\begin{array}{l} \cos^{’ 1} (\cos \frac{13}{6}) „ = „ \frac{13}{6} \\ But \frac{13}{6} „ ˆ‰ „ [0 „] \\ \cos „ (\frac{13}{6}) „ = \cos (2 + \frac{}{6}) „ = „ \cos „ (\frac{}{6}) \\ ˆ´ „ „ \cos^{1} „ (\cos \frac{13}{6}) „ = „ \frac{}{6} \end{array}$

Q.4

$\begin{array}{l} Find the principal value of \\ (i) {sin}^{-1} (\frac{-1}{2}) \\ (ii) {cot}^{-1} (\frac{-1}{\sqrt{3}}) \end{array}$

Option –

$\begin{array}{l} (i) \sin^{- 1} (\frac{- 1}{2}) = y \\ \sin y = \frac{1}{2} \\ \sin y = \sin \frac{}{} \end{array}$

Ans.

$\begin{array}{l} (i) \sin^{- 1} (\frac{- 1}{2}) = y \\ \sin y = \frac{1}{2} \\ \sin y = \sin \frac{}{6} \\ \sin y = \sin (\frac{}{6}) or y = \frac{}{6} \\ We know that the range of inverse sine function is [\frac{}{2} \frac{}{2}] \\ ˆ´ \sin^{1} (\frac{}{6}) = \frac{}{6} \\ (ii) Let \cot^{1} (\frac{1}{\sqrt{3}}) = y \\ coty = \frac{1}{\sqrt{3}} \\ coty = \frac{1}{\sqrt{3}} \\ coty = \cot (\frac{}{3}) \\ y = \frac{2}{3} \\ We know that range of inverse contangent function (0) \\ ˆ´ \cot^{1} ‹ (\frac{1}{\sqrt{3}}) = \frac{2}{3} \end{array}$

Q5-

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

Opt-

$\begin{array}{l} \tan^{1} (1) + \cos^{- 1} (\frac{- 1}{2}) + \sin^{- 1} (\frac{- 1}{2}) \\ \tan^{1} (\tan \frac{}{4}) + \cos^{- 1} (\cos \frac{}{3}) + sin< \end{array}$

Ans-

$\begin{array}{l} \tan^{1} (1) + \cos^{- 1} (\frac{- 1}{2}) + \sin^{- 1} (\frac{- 1}{2}) \\ \tan^{1} (\tan \frac{}{4}) + \cos^{- 1} (\cos \frac{}{3}) + \sin^{- 1} (\sin \frac{}{6}) \\ = \frac{}{4} + \cos^{- 1} (\cos (\frac{}{3})) + \sin^{- 1} [\sin (\frac{}{6})] \\ we know range of \\ \tan^{1} †’ (\frac{’}{2} \frac{}{2}) \\ \cos^{1} †’ [0] and \sin^{1} †’ [\frac{}{2} \frac{}{2}] \\ \tan^{ˆ 1} (1) + \cos^{1} (\frac{’ 1}{2}) + \sin^{1} (\frac{1}{2}) \\ = \frac{}{4} + \frac{2}{3} \frac{}{6} \\ = \frac{3 + 8 ’ 2}{12} = \frac{9}{12} = \frac{3}{4} \end{array}$

CBSE Important Questions Class 12 Maths Chapter 2 – Inverse Trigonometric Functions

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