Inverse Trigonometric Functions — Class 12 Mathematics NCERT Solutions (Free)

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TL;DR: Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Inverse Trigonometric Functions" — 8 important questions with detailed answers for…

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NCERT Solutions for Class 12 Mathematics — Inverse Trigonometric Functions. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.

Inverse Trigonometric Functions — All 8 Questions Solved

Q1. Find the principal value of sin⁻¹(-1/2).

We need to find the principal value of sin⁻¹(-1/2).

Let θ = sin⁻¹(-1/2)
Then sin(θ) = -1/2

The range of sin⁻¹ is [-π/2, π/2].

We know sin(π/6) = 1/2.
Since we need a negative value and the principal range is [-π/2, π/2]:
sin(-π/6) = -sin(π/6) = -1/2

Since -π/6 ∈ [-π/2, π/2], this is the principal value.

CONCLUSION: sin⁻¹(-1/2) = -π/6

Q2. Prove that tan⁻¹(x) + tan⁻¹(1/x) = π/2 for x > 0.

We need to prove: tan⁻¹(x) + tan⁻¹(1/x) = π/2 for x > 0

Let α = tan⁻¹(x) and β = tan⁻¹(1/x), where x > 0.

Then tan(α) = x and tan(β) = 1/x
Also, 0 < α < π/2 and 0 < β < π/2 (since x > 0)

We compute tan(α + β):
tan(α + β) = (tan(α) + tan(β))/(1 - tan(α)tan(β))

Substituting:
tan(α + β) = (x + 1/x)/(1 - x(1/x))
= (x + 1/x)/(1 - 1)
= (x + 1/x)/0
= undefined

tan(α + β) is undefined, which means α + β = π/2 (since α + β ∈ (0, π)).

Therefore, tan⁻¹(x) + tan⁻¹(1/x) = π/2 for x > 0. ✓

Q3. Find the value of cos⁻¹(cos(7π/6)).

We need to find cos⁻¹(cos(7π/6)).

The range of cos⁻¹ is [0, π].

First, let's find cos(7π/6):
7π/6 = π + π/6
cos(7π/6) = cos(π + π/6) = -cos(π/6) = -√3/2

Now we need to find cos⁻¹(-√3/2).
This means finding θ ∈ [0, π] such that cos(θ) = -√3/2.

We know cos(π/6) = √3/2.
For the negative value in the range [0, π]:
cos(π - π/6) = cos(5π/6) = -cos(π/6) = -√3/2

Since 5π/6 ∈ [0, π], we have:
cos⁻¹(-√3/2) = 5π/6

CONCLUSION: cos⁻¹(cos(7π/6)) = 5π/6

Q4. Find the principal value of tan⁻¹(√3).

We need to find the principal value of tan⁻¹(√3).

Let θ = tan⁻¹(√3)
Then tan(θ) = √3

The range of tan⁻¹ is (-π/2, π/2).

We know that tan(π/3) = √3.

Since π/3 ∈ (-π/2, π/2), we have:

CONCLUSION: tan⁻¹(√3) = π/3

Q5. Simplify: sin⁻¹(3/5) + sin⁻¹(4/5).

Let α = sin⁻¹(3/5) and β = sin⁻¹(4/5)

Then sin(α) = 3/5 and sin(β) = 4/5, where α, β ∈ [0, π/2] (since both values are positive)

Finding cos(α) and cos(β):
cos(α) = √(1 - sin²(α)) = √(1 - 9/25) = √(16/25) = 4/5 (positive in first quadrant)
cos(β) = √(1 - sin²(β)) = √(1 - 16/25) = √(9/25) = 3/5 (positive in first quadrant)

Finding sin(α + β):
sin(α + β) = sin(α)cos(β) + cos(α)sin(β)
= (3/5)(3/5) + (4/5)(4/5)
= 9/25 + 16/25
= 25/25
= 1

Since sin(α + β) = 1 and α + β ∈ [0, π], we have:
α + β = π/2

Therefore: sin⁻¹(3/5) + sin⁻¹(4/5) = π/2

CONCLUSION: The sum equals π/2

Q6. If sin⁻¹(x) = π/6, find the value of x.

Given: sin⁻¹(x) = π/6

By definition of inverse sine function:
if sin⁻¹(x) = π/6, then sin(π/6) = x

We know that sin(π/6) = 1/2

Therefore: x = 1/2

VERIFICATION:
sin⁻¹(1/2) = π/6 ✓ (since sin(π/6) = 1/2 and π/6 is in the range [-π/2, π/2])

CONCLUSION: x = 1/2

Q7. Find the principal value of sec⁻¹(-2).

We need to find the principal value of sec⁻¹(-2).

Let θ = sec⁻¹(-2)
Then sec(θ) = -2, which means cos(θ) = -1/2

The range of sec⁻¹ is [0, π] \ {π/2}.

We know cos(π/3) = 1/2.
For cos(θ) = -1/2 in the range [0, π]:
θ = π - π/3 = 2π/3

Verification: cos(2π/3) = -cos(π/3) = -1/2 ✓
sec(2π/3) = 1/cos(2π/3) = 1/(-1/2) = -2 ✓

CONCLUSION: sec⁻¹(-2) = 2π/3

Q8. Evaluate: cos⁻¹(1/2) + 2sin⁻¹(1/2).

We need to evaluate cos⁻¹(1/2) + 2sin⁻¹(1/2).

Finding cos⁻¹(1/2):
Let θ = cos⁻¹(1/2)
Then cos(θ) = 1/2, and θ ∈ [0, π]
Since cos(π/3) = 1/2, we have cos⁻¹(1/2) = π/3

Finding sin⁻¹(1/2):
Let φ = sin⁻¹(1/2)
Then sin(φ) = 1/2, and φ ∈ [-π/2, π/2]
Since sin(π/6) = 1/2, we have sin⁻¹(1/2) = π/6

Evaluating the expression:
cos⁻¹(1/2) + 2sin⁻¹(1/2) = π/3 + 2(π/6)
= π/3 + π/3
= 2π/3

CONCLUSION: cos⁻¹(1/2) + 2sin⁻¹(1/2) = 2π/3

Inverse Trigonometric Functions — Practice MCQs with Answers

Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.

MCQ 1. The principal value of $\sin^{-1}\left(-\frac{1}{2}\right)$ is:

• A. $\frac{\pi}{6}$
• B. $-\frac{\pi}{6}$
• C. $\frac{5\pi}{6}$
• D. $-\frac{\pi}{3}$

Answer: B. $-\frac{\pi}{6}$ — The principal value branch of $\sin^{-1}x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Since $\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}$, the principal value is $-\frac{\pi}{6}$.

MCQ 2. The domain of the function $f(x) = \cos^{-1}(2x-1)$ is:

• A. $[0, 1]$
• B. $[-1, 1]$
• C. $[0, \pi]$
• D. $[-\frac{\pi}{2}, \frac{\pi}{2}]$

Answer: A. $[0, 1]$ — The domain of $\cos^{-1}y$ is $[-1, 1]$. Therefore, $-1 \le 2x-1 \le 1$. Adding 1 to all parts, we get $0 \le 2x \le 2$, which simplifies to $0 \le x \le 1$.

MCQ 3. The value of $\sin\left(\sin^{-1}\left(\frac{1}{2}\right) + \cos^{-1}\left(\frac{1}{2}\right)\right)$ is:

• A. $0$
• B. $1$
• C. $\frac{1}{2}$
• D. $\frac{\sqrt{3}}{2}$

Answer: B. $1$ — We know that $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ for $x \in [-1, 1]$. Here $x = \frac{1}{2}$, so the expression becomes $\sin\left(\frac{\pi}{2}\right)$, which is $1$.

MCQ 4. If $\tan^{-1}x = y$, then $\cos y$ is equal to:

• A. $\frac{1}{\sqrt{1+x^2}}$
• B. $\frac{x}{\sqrt{1+x^2}}$
• C. $\sqrt{1+x^2}$
• D. $\frac{1}{x}$

Answer: A. $\frac{1}{\sqrt{1+x^2}}$ — If $\tan^{-1}x = y$, then $\tan y = x$. Consider a right-angled triangle with opposite side $x$ and adjacent side $1$. The hypotenuse is $\sqrt{x^2+1^2}$. Then $\cos y = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{1+x^2}}$.

MCQ 5. The value of $\tan^{-1}\left(\frac{1}{2} ight) + \tan^{-1}\left(\frac{1}{3} ight)$ is:

• A. $\frac{\pi}{4}$
• B. $\frac{\pi}{2}$
• C. $\frac{\pi}{3}$
• D. $\frac{3\pi}{4}$

Answer: A. $\frac{\pi}{4}$ — Using the formula $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ for $xy < 1$. Here $x=\frac{1}{2}, y=\frac{1}{3}$, so $xy=\frac{1}{6} < 1$. $\tan^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot\frac{1}{3}}\right) = \tan^{-1}\left(\frac{\frac{5}{6}}{1-\frac{1}{6}}\right) = \tan^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right) = \tan^{-1}(1) = \frac{\pi}{4}$.

MCQ 6. The value of $\sin^{-1}\left(\sin\left(\frac{3\pi}{5}\right)\right)$ is:

• A. $\frac{3\pi}{5}$
• B. $\frac{2\pi}{5}$
• C. $-\frac{3\pi}{5}$
• D. $\frac{\pi}{5}$

Answer: B. $\frac{2\pi}{5}$ — The principal value branch of $\sin^{-1}x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Since $\frac{3\pi}{5}$ is not in this interval, we write $\sin\left(\frac{3\pi}{5} ight) = \sin\left(\pi - \frac{3\pi}{5} ight) = \sin\left(\frac{2\pi}{5} ight)$. Since $\frac{2\pi}{5} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, we have $\sin^{-1}\left(\sin\left(\frac{3\pi}{5} ight)\right) = \sin^{-1}\left(\sin\left(\frac{2\pi}{5} ight)\right) = \frac{2\pi}{5}$.

MCQ 7. The principal value of $\tan^{-1}(-1)$ is:

• A. $\frac{\pi}{4}$
• B. $\frac{3\pi}{4}$
• C. $-\frac{\pi}{4}$
• D. $-\frac{3\pi}{4}$

Answer: C. $-\frac{\pi}{4}$ — The principal value branch of $\tan^{-1}x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$. Since $\tan\left(-\frac{\pi}{4}\right) = -1$, the principal value is $-\frac{\pi}{4}$.

MCQ 8. Simplify $\tan^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}\right)$, where $x \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$.

• A. $\frac{\pi}{4} - x$
• B. $\frac{\pi}{4} + x$
• C. $x - \frac{\pi}{4}$
• D. $\frac{\pi}{2} - x$

Answer: A. $\frac{\pi}{4} - x$ — Divide the numerator and denominator by $\cos x$: $\tan^{-1}\left(\frac{1 - \tan x}{1 + \tan x}\right)$. This is the formula for $\tan\left(\frac{\pi}{4} - x\right)$. Since $x \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$, then $\frac{\pi}{4}-x \in (0, \frac{\pi}{2})$, which is within the principal branch of $\tan^{-1}$. Thus, the value is $\frac{\pi}{4} - x$.

MCQ 9. The value of $\sin^{-1}x + \cos^{-1}x$ for $x \in [-1, 1]$ is:

• A. $\pi$
• B. $\frac{\pi}{2}$
• C. $0$
• D. $2\pi$

Answer: B. $\frac{\pi}{2}$ — This is a fundamental property of inverse trigonometric functions, stating that $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ for all $x$ in their common domain $[-1, 1]$.

MCQ 10. If $\sin^{-1}x + \sin^{-1}y = \frac{2\pi}{3}$, then $\cos^{-1}x + \cos^{-1}y$ is equal to:

• A. $\frac{\pi}{3}$
• B. $\frac{2\pi}{3}$
• C. $\pi$
• D. $\frac{4\pi}{3}$

Answer: A. $\frac{\pi}{3}$ — We know that $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ and $\sin^{-1}y + \cos^{-1}y = \frac{\pi}{2}$. Adding these two identities, we get $(\sin^{-1}x + \sin^{-1}y) + (\cos^{-1}x + \cos^{-1}y) = \pi$. Substituting the given value, $\frac{2\pi}{3} + (\cos^{-1}x + \cos^{-1}y) = \pi$. Therefore, $\cos^{-1}x + \cos^{-1}y = \pi - \frac{2\pi}{3} = \frac{\pi}{3}$.

MCQ 11. The value of $2\tan^{-1}\left(\frac{1}{3} ight) + \tan^{-1}\left(\frac{1}{7} ight)$ is:

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{4}$
• C. $\tan^{-1}\left(\frac{31}{17} ight)$
• D. $\pi$

Answer: B. $\frac{\pi}{4}$ — First, use the formula $2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$. So, $2\tan^{-1}\left(\frac{1}{3} ight) = \tan^{-1}\left(\frac{2 \cdot \frac{1}{3}}{1 - \left(\frac{1}{3}\right)^2}\right) = \tan^{-1}\left(\frac{\frac{2}{3}}{\frac{8}{9}}\right) = \tan^{-1}\left(\frac{3}{4} ight)$. Now, use $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy} ight)$: $\tan^{-1}\left(\frac{3}{4}\right) + \tan^{-1}\left(\frac{1}{7}\right) = \tan^{-1}\left(\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4}\cdot\frac{1}{7}}\right) = \tan^{-1}\left(\frac{\frac{25}{28}}{\frac{25}{28}}\right) = \tan^{-1}(1) = \frac{\pi}{4}$.

MCQ 12. The principal value of $\cos^{-1}\left(-\frac{1}{2} ight)$ is:

• A. $\frac{\pi}{3}$
• B. $-\frac{\pi}{3}$
• C. $\frac{2\pi}{3}$
• D. $\frac{4\pi}{3}$

Answer: C. $\frac{2\pi}{3}$ — The principal value branch of $\cos^{-1}x$ is $[0, \pi]$. We know $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$. Since $\cos^{-1}(-x) = \pi - \cos^{-1}x$, then $\cos^{-1}\left(-\frac{1}{2}\right) = \pi - \cos^{-1}\left(\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

MCQ 13. The value of $\tan^{-1}\sqrt{3} - \sec^{-1}(-2)$ is:

• A. $\frac{\pi}{3}$
• B. $-\frac{\pi}{3}$
• C. $\frac{\pi}{6}$
• D. $-\frac{\pi}{6}$

Answer: B. $-\frac{\pi}{3}$ — We know $\tan^{-1}\sqrt{3} = \frac{\pi}{3}$. For $\sec^{-1}(-2)$, let $y = \sec^{-1}(-2)$. Then $\sec y = -2$. The principal value branch for $\sec^{-1}x$ is $[0, \pi] - \{\frac{\pi}{2}\}$. We know $\sec(\frac{\pi}{3}) = 2$. So $\sec y = -\sec(\frac{\pi}{3}) = \sec(\pi - \frac{\pi}{3}) = \sec(\frac{2\pi}{3})$. Thus, $\sec^{-1}(-2) = \frac{2\pi}{3}$. Therefore, $\tan^{-1}\sqrt{3} - \sec^{-1}(-2) = \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}$.

MCQ 14. The solution of the equation $\tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4}$ is:

• A. $\frac{1}{6}$
• B. $-\frac{1}{6}$
• C. $1$
• D. $\frac{1}{3}$

Answer: A. $\frac{1}{6}$ — Using $\tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$, we get $\frac{5x}{1-6x^2} = 1$, which leads to $6x^2+5x-1=0$. Factoring gives $(6x-1)(x+1)=0$, so $x=\frac{1}{6}$ or $x=-1$. Checking these solutions, for $x=-1$, $2x=-2, 3x=-3$, so $xy = 6 > 1$. In this case, $\tan^{-1}(-2) + \tan^{-1}(-3) = -\pi + \tan^{-1}\left(\frac{-2-3}{1-6}\right) = -\pi + \tan^{-1}(1) = -\frac{3\pi}{4} \neq \frac{\pi}{4}$. For $x=\frac{1}{6}$, $2x=\frac{1}{3}, 3x=\frac{1}{2}$, so $xy = \frac{1}{6} < 1$, the formula applies directly: $\tan^{-1}\left(\frac{1/3+1/2}{1-1/6}\right) = \tan^{-1}(1) = \frac{\pi}{4}$. Thus, $x=\frac{1}{6}$ is the only valid solution.

MCQ 15. The value of $\tan(\tan^{-1}(1))$ is:

• A. $0$
• B. $1$
• C. $\infty$
• D. Undefined

Answer: B. $1$ — The property $\tan(\tan^{-1}x) = x$ holds for all real $x$. Here $x=1$, so the value is $1$.

MCQ 16. The range of the function $f(x) = \sin^{-1}x$ is:

• A. $(-\infty, \infty)$
• B. $[0, \pi]$
• C. $[-\frac{\pi}{2}, \frac{\pi}{2}]$
• D. $[0, \frac{\pi}{2}]$

Answer: C. $[-\frac{\pi}{2}, \frac{\pi}{2}]$ — By definition, the principal value branch (range) of the inverse sine function, $\sin^{-1}x$, is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

MCQ 17. $\sin^{-1}\left(\frac{3}{5}\right)$ is equal to:

• A. $\tan^{-1}\left(\frac{3}{4}\right)$
• B. $\tan^{-1}\left(\frac{4}{3}\right)$
• C. $\tan^{-1}\left(\frac{5}{3}\right)$
• D. $\tan^{-1}\left(\frac{5}{4}\right)$

Answer: A. $\tan^{-1}\left(\frac{3}{4}\right)$ — Let $\theta = \sin^{-1}\left(\frac{3}{5}\right)$. Then $\sin\theta = \frac{3}{5}$. Consider a right-angled triangle where the opposite side is $3$ and the hypotenuse is $5$. The adjacent side is $\sqrt{5^2 - 3^2} = 4$. Thus, $\tan\theta = \frac{3}{4}$. So $\theta = \tan^{-1}\left(\frac{3}{4}\right)$.

MCQ 18. The value of $\cos^{-1}\left(\cos\left(\frac{7\pi}{6}\right)\right)$ is:

• A. $\frac{7\pi}{6}$
• B. $\frac{5\pi}{6}$
• C. $\frac{\pi}{6}$
• D. $-\frac{\pi}{6}$

Answer: B. $\frac{5\pi}{6}$ — The principal value branch of $\cos^{-1}x$ is $[0, \pi]$. Since $\frac{7\pi}{6}$ is not in this interval, we use the property $\cos x = \cos(2\pi - x)$. So, $\cos\left(\frac{7\pi}{6}\right) = \cos\left(2\pi - \frac{7\pi}{6} ight) = \cos\left(\frac{5\pi}{6}\right)$. Since $\frac{5\pi}{6} \in [0, \pi]$, we have $\cos^{-1}\left(\cos\left(\frac{7\pi}{6} ight)\right) = \cos^{-1}\left(\cos\left(\frac{5\pi}{6} ight)\right) = \frac{5\pi}{6}$.

MCQ 19. If $\operatorname{cosec}^{-1}x = y$, then $\operatorname{cosec}^{-1}(-x)$ is equal to:

• A. $-y$
• B. $\pi - y$
• C. $\pi + y$
• D. $\frac{\pi}{2} - y$

Answer: A. $-y$ — The property for $\operatorname{cosec}^{-1}(-x)$ is $\operatorname{cosec}^{-1}(-x) = -\operatorname{cosec}^{-1}x$. So, if $\operatorname{cosec}^{-1}x = y$, then $\operatorname{cosec}^{-1}(-x) = -y$.

MCQ 20. Simplify $\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$, for $x \neq 0$.

• A. $\tan^{-1}x$
• B. $\frac{1}{2}\tan^{-1}x$
• C. $2\tan^{-1}x$
• D. $\frac{\pi}{2} - \tan^{-1}x$

Answer: B. $\frac{1}{2}\tan^{-1}x$ — Let $x = \tan\theta$, so $\theta = \tan^{-1}x$. The expression becomes $\tan^{-1}\left(\frac{\sec\theta-1}{\tan\theta}\right) = \tan^{-1}\left(\frac{1-\cos\theta}{\sin\theta}\right)$. Using half-angle formulas, this simplifies to $\tan^{-1}\left(\frac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)}\right) = \tan^{-1}\left(\tan\left(\frac{\theta}{2}\right)\right) = \frac{\theta}{2}$. Substituting back $\theta = \tan^{-1}x$, the simplified expression is $\frac{1}{2}\tan^{-1}x$.

How to Use These Solutions

Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.

Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.

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