Inverse Trigonometric Functions

Previous Years' CBSE Board Questions

2.2 Basic Concepts

MCQ

1. <math>\sin \left[ \frac{\pi}{3} + \sin^{-1} \left( \frac{1}{2} \right) \right]</math> is equal to (a) 1 (b) <math>\frac{1}{2}</math> (c) <math>\frac{1}{3}</math> (d) <math>\frac{1}{4}</math> (2023) 11. Write the principal value of <math>\tan^{-1}\left[\sin\left(-\frac{\pi}{2}\right)\right]</math>.

1. If <math>f(x) = |\cos x|</math>, then <math>f\left(\frac{3\pi}{4}\right)</math> is (a) 1 (b) -1 (c) <math>\frac{-1}{\sqrt{2}}</math> (d) <math>\frac{1}{\sqrt{2}}</math> (2023)
2. Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (a), (b), (c) and (d) as given below.

Assertion (A): All trigonometric functions have their inverses over their respective domains. Reason (R): The inverse of <math>tan^{-1}x</math> exists for some <math>x \in \mathbb{R}</math>.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

(2020) Ev

1. Assertion (A) is true but Reason (R) is false.
2. Assertion (A) is false but Reason (R) is true.

(2023)

1. The value of <math>\sin^{-1}\left(\cos\frac{13\pi}{5}\right)</math> is
2. <math>-\frac{3\pi}{5}</math> (b) <math>-\frac{\pi}{10}</math> (c) <math>\frac{3\pi}{5}</math> (d) <math>\frac{\pi}{10}</math> (Term I, 2021-22) 🕕

1. The principal value of <math>\cot^{-1}(-\sqrt{3})</math> is
2. <math>-\frac{\pi}{6}</math> (b) <math>\frac{\pi}{6}</math> (c) <math>\frac{2\pi}{3}</math> (d) <math>\frac{5\pi}{6}</math>

(2020)

(2020, 2019 C)

1. <math>\tan^{-1} 3 + \tan^{-1} \lambda = \tan^{-1} \left( \frac{3+\lambda}{1-3\lambda} \right)</math> is valid for what values of <math>\lambda</math>?
2. <math>\lambda \in \left(-\frac{1}{3}, \frac{1}{3}\right)</math> (b) <math>\lambda > \frac{1}{3}</math>
3. <math>\lambda < \frac{1}{2}</math> (d) All real values of λ

(2019)

1. The principal value of <math>tan^{-1}\left(tan\frac{3\pi}{5}\right)</math> is
2. <math>\frac{2\pi}{5}</math> (b) <math>\frac{-2\pi}{5}</math> (c) <math>\frac{3\pi}{5}</math> (d) <math>\frac{-3\pi}{5}</math>

(2020) U

V5A (1 mark)

The range of the principal value branch of the function <math>y = \sec^{-1}x</math> is _____. (2020)

1. The principal value of <math>\cos^{-1}\left(\frac{-1}{2}\right)</math> is ______. (2020)
2. Write the value of <math>\cos^{-1}\left(-\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right)</math> (Foreign 2014) An

(AI 2014C)

12. Find the value of the following:

<math>\cot\left(\frac{\pi}{2}-2\cot^{-1}\sqrt{3}\right)</math> (AI 2014C)

(2 marks)

Write the domain and range (principle value branch)

of the following functions:

<math>f(x) = \tan^{-1}x</math>. (2023)

1. Evaluate: <math>\cos^{-1} \left| \cos \left( -\frac{7\pi}{2} \right) \right|</math> (2023)
2. Simplify <math>\sec^{-1}\left(\frac{1}{2x^2+1}\right)0 < x < \frac{1}{\sqrt{2}}</math>. (2021C)
3. Prove that: <math>\frac{9\pi}{8} - \frac{9}{4} \sin^{-1} \left( \frac{1}{3} \right) = \frac{9}{4} \sin^{-1} \left( \frac{2\sqrt{2}}{3} \right)</math>
4. Prove that: <math>\sin^{-1}(2x\sqrt{1-x^2}) = 2\cos^{-1}x, \frac{1}{\sqrt{2}} \le x \le 1.</math> (2020)

PA (4 marks)

1. Solve for x: <math>\sin^{-1}(1-x)-2\sin^{-1}x=\frac{\pi}{2}</math> (2020C)
2. Prove that: <math>2\tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{7}=\tan^{-1}\frac{31}{47}</math>. (2020C)
3. Prove that: <math>\tan^{-1} \sqrt{x} = \frac{1}{2} \cos^{-1} \left( \frac{1-x}{1+x} \right) x \in [0,1]</math>
4. Prove that: <math>\cos^{-1}\left(\frac{12}{13}\right) + \sin^{-1}\left(\frac{3}{5}\right) = \sin^{-1}\left(\frac{56}{45}\right)</math> (Al 2019) EV
5. Prove that: <math>\sin^{-1}\frac{4}{5} + \tan^{-1}\frac{5}{12} + \cos^{-1}\frac{63}{45} = \frac{\pi}{2}</math>
6. If <math>tan^{-1}x - cot^{-1}x = tan^{-1}\left(\frac{1}{\sqrt{3}}\right)</math> x > 0, find the value of x and hence find the value of <math>\sec^{-1}\left(\frac{2}{x}\right)</math> (2019) [1]
7. Find the value of <math>\sin\left(\cos^{-1}\frac{4}{5} + \tan^{-1}\frac{2}{3}\right)</math> (2019)
8. Prove that : <math>\tan^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x; -\frac{1}{\sqrt{2}} \le x \le 1</math> (2019C) Ap

CBSE Sample Questions

2.2 Basic Concepts

MCQ

In the given question, a statement of assertion (A) is followed by a statement of reason (R). Choose the correct answer out of the following choices. Assertion (A): The domain of the function sec-1 2x is <math>\left(-\infty, -\frac{1}{2}\right] \cup \left[\frac{1}{2}, \infty\right)</math>

<math>\tan^{-1}\left(\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}\right)\pi < x < \frac{3\pi}{2}</math> is

(a) <math>\frac{\pi}{4} - \frac{x}{2}</math> (b) <math>\frac{3\pi}{2} - \frac{x}{2}</math> (c) <math>-\frac{x}{2}</math> (d) <math>\pi - \frac{x}{2}</math>

(Term I, 2021-22) Cr 5.

Reason (R): <math>sec^{-1}(-2) = -\frac{\pi}{4}</math>

1. Both A and R are true and R is the correct explanation of A.

Both A and R are true but R is not the correct explanation of A.

1. A is true but R is false.

(2022-23) (c) <math>\frac{-\pi}{2} < y < \frac{\pi}{2}</math> (d) <math>y \in \left\{ \frac{-\pi}{2}, \frac{\pi}{2} \right\}</math>

1. A is false but R is true.

1. <math>\sin \left[ \frac{\pi}{3} - \sin^{-1} \left( -\frac{1}{2} \right) \right]</math> is equal to
2. <math>\frac{1}{2}</math> (b) <math>\frac{1}{2}</math> (c) -1 (d) 1 (Term I, 2021-22) (Aii)

(2022-23)

1. <math>\sin (\tan^{-1}x)</math>, where <math>|x| < 1</math>, is equal to
2. <math>\frac{x}{1 + 2}</math> (b) <math>\frac{1}{1 + 2}</math>

1. <math>\frac{1}{\sqrt{1+x^2}}</math>
2. <math>\frac{x}{\sqrt{1+x^2}}</math>

(Term I, 2021-22)

Simplest form of 4.

If <math>tan^{-1}x = y</math>, then

1. -1 < y < 1

(b) <math>\frac{-\pi}{2} \le y \le \frac{\pi}{2}</math>

(Term I, 2021-22)

SAI (2 marks)

1. Find the value of <math>\sin^{-1} \left[ \sin \left( \frac{13\pi}{7} \right) \right]</math>.
2. Express <math>\tan^{-1} \left( \frac{\cos x}{1 - \sin x} \right) \frac{-3\pi}{2} < x < \frac{\pi}{2}</math> in the simplest form. (2020-21) EV

Detailed SOLUTIONS

1 (a): We have,

<math>\sin\left|\frac{\pi}{3} + \sin^{-1}\left(\frac{1}{2}\right)\right| = \sin\left|\frac{\pi}{3} + \sin^{-1}\left(\sin\frac{\pi}{6}\right)\right| = \sin\left|\frac{\pi}{3} + \frac{\pi}{6}\right|</math> <math>=\sin\left(\frac{\pi}{2}\right)=1</math>

(d): f(x) = |cosx|

At <math>\frac{\pi}{2} < x < \pi</math>, <math>\cos x < 0</math> : <math>|\cos x| = -\cos x \Rightarrow f(x) = -\cos x</math>

<math display="block">\therefore f\left(\frac{3\pi}{4}\right) = -\cos\left(\frac{3\pi}{4}\right) = -\cos\left(\pi - \frac{\pi}{4}\right)</math> <math>=\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}</math> [:: <math>cos(\pi - \theta) = -cos\theta</math>] 5. (d): We know that <math>\cot^{-1}(x) \in (0, \pi)</math>

1. (d): All trigonometric functions are periodic and

hence not invertible over their respective domains but all trigonometric functions have inverse over their restricted domains.

<math>=\cot^{-1}\left|\cot\left(\pi-\frac{\pi}{4}\right)\right|</math> <math display="block">[\because \cot(\pi - \theta) = -\cot\theta]</math>

Inverse of tan-1x is tanx which is defined for

<math>\left[ \because \cot^{-1}[\cot\theta] = \theta \right]</math>

<math>x \in R - (2n+1)\frac{\pi}{2}, n \in Z</math>

7 Assertion is false and reason is true.

4. (b): We have, <math>\sin^{-1}\left(\cos\frac{13\pi}{5}\right) = \sin^{-1}\left[\cos\left(2\pi + \frac{3\pi}{5}\right)\right]</math>

<math display="block">=\sin^{-1}\left[\cos\frac{3\pi}{5}\right] = \sin^{-1}\left[\cos\left(\frac{\pi}{2} + \frac{\pi}{10}\right)\right]</math>

<math>=\sin^{-1}\left(-\sin\frac{\pi}{10}\right) = -\sin^{-1}\left(\sin\frac{\pi}{10}\right) = -\frac{\pi}{10}</math>

Answer Tips (💅

<math>\Rightarrow</math> <math>\cos(2\pi + \theta) = \cos\theta, \cos\left(\frac{\pi}{2} + \theta\right) = -\sin\theta</math>

<math>\cot^{-1}(-\sqrt{3}) = \cot^{-1}(-\cot\frac{\pi}{6})</math>

<math>=\cot^{-1}\left|\cot\left(\frac{5\pi}{6}\right)\right| = \frac{5\pi}{6}</math>

Thus, the principal value of <math>\cot^{-1}(-\sqrt{3})</math> is <math>\frac{5\pi}{4}</math>.