CBSE Class 12 Mathematics Inverse Trigonometric Functions Class Test Set 01

This Class 12 CBSE Mathematics class test on Chapter 2, Inverse Trigonometric Functions, has 11 questions carrying 1 to 3 marks each. Full step-by-step solutions are given for every question.

Class Test Questions with Solutions

Question 1. Find the principal value of \( \tan^{-1}\left\{\sin\left(-\frac{\pi}{2}\right)\right\} \) [1 Mark]
Answer:
We know \( \sin\left(\frac{-\pi}{2}\right) = -1 \)
\( \therefore \tan^{-1}\left\{\sin \frac{-\pi}{2}\right\} \)
\( = \tan^{-1}(-1) = \frac{-\pi}{4} \)

Question 2. Which is greater, tan 1 or \( \tan^{-1} 1 \)? [1 Mark]
Answer:
We know \( \tan^{-1} 1 = \frac{\pi}{4} \)
Now \( 1 > \frac{\pi}{4} \)
\( \tan 1 > \tan \frac{\pi}{4} \)
tan 1 > 1
\( \therefore \tan 1 > \frac{\pi}{4} \)
\( \Rightarrow \tan 1 > \tan^{-1} 1 \)

Question 3. Find the domain of \( f(x) = \sin^{-1}(-x^2) \). [1 Mark]
Answer:
The domain of \( \sin^{-1} x \) is [−1, 1]
\( \therefore -1 \le -x^2 \le 1 \)
\( \Rightarrow 0 \le x^2 \le 1 \)
\( \Rightarrow x^2 \le 1 \)
\( \Rightarrow -1 \le x \le 1 \)
\( \therefore \) Domain of \( \sin^{-1}(-x^2) \) is [−1, 1]

Question 4. Find the domain of \( f(x) = \sin^{-1} x + \cos x \). [1 Mark]
Answer:
Domain of \( \sin^{-1} x \) is [−1, 1] and cos x is R
\( \therefore \) Domain of \( f(x) = \sin^{-1} x + \cos x \) is
[−1, 1] ∩ R
= [−1, 1]

Question 5. Find the principal value of \( \cos^{-1}\left\{\sin\left(\cos^{-1} \frac{1}{2}\right)\right\} \) [2 Marks]
Answer:
We know that \( \cos^{-1} \frac{1}{2} = \frac{\pi}{3} \)
\( \therefore \cos^{-1}\left\{\sin\left(\cos^{-1} \frac{1}{2}\right)\right\} \)
\( = \cos^{-1}\left(\sin \frac{\pi}{3}\right) = \cos^{-1} \frac{\sqrt{3}}{2} \)
\( = \frac{\pi}{6} \)

Question 6. Find the value of \( \sin^{-1}\left[\cos\left\{\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right\}\right] \) [2 Marks]
Answer:
\( \sin^{-1}\left[\cos\left\{\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right\}\right] \)
\( = \sin^{-1}\left\{\cos\left(\frac{-\pi}{3}\right)\right\} \)
\( = \sin^{-1}\left(\cos \frac{\pi}{3}\right) \)
\( = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \)

Question 7. For the principal values, evaluate the following: [3 Marks]
(i) \( \cot^{-1}(-1) + \text{cosec}^{-1}(-\sqrt{2}) + \sec^{-1}(2) \)
(ii) \( \cot^{-1}(-\sqrt{3}) + \tan^{-1}(1) + \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) \)
Answer:
(i) \( \cot^{-1}(-1) + \text{cosec}^{-1}(-\sqrt{2}) + \sec^{-1}(2) \)
\( = \frac{3\pi}{4} - \frac{\pi}{4} + \frac{\pi}{3} = \frac{5\pi}{6} \)
(ii) \( \cot^{-1}(-\sqrt{3}) + \tan^{-1}(1) + \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) \)
\( = \frac{5\pi}{6} + \frac{\pi}{4} + \frac{\pi}{6} = \frac{5\pi}{4} \)

Question 8. (i) Find the domain of \( \sec^{-1}(2x + 1) \) [3 Marks]
(ii) Find the principal values of \( \sec^{-1} \frac{2}{\sqrt{3}} \) and \( \sec^{-1}(-2) \).
Answer:
(i) Domain of \( \sec^{-1} x \) is (−∞, −1] ∪ [1, ∞)
\( \therefore \sec^{-1}(2x + 1) \) is meaningful, if
2x + 1 ≥ 1 or 2x + 1 ≤ −1
x ≥ 0 or x ≤ −1
\( \therefore \) x ∈ (−∞, −1] ∪ [0, ∞)
(ii) Let \( \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \theta \)
\( \theta = \frac{\pi}{6} \)
Also let \( \sec^{-1}(-2) = \phi \)
\( \phi = \frac{2\pi}{3} \)

Question 9. For the principal values, evaluate each of the following: [3 Marks]
(i) \( \tan^{-1}\left\{2\cos\left(2\sin^{-1} \frac{1}{2}\right)\right\} \) (ii) \( \cot\left[\sin^{-1}\left\{\cos(\tan^{-1} 1)\right\}\right] \)
Answer:
(i) We know \( \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \)
\( \therefore \tan^{-1}\left\{2\cos\left(2\sin^{-1} \frac{1}{2}\right)\right\} \)
\( = \tan^{-1}\left\{2\cos\left(2 \times \frac{\pi}{6}\right)\right\} \)
\( = \tan^{-1}\left(2\cos \frac{\pi}{3}\right) \)
\( = \tan^{-1}\left(2 \times \frac{1}{2}\right) = \frac{\pi}{4} \)
(ii) We know \( \tan^{-1} 1 = \frac{\pi}{4} \)
\( \cot\left[\sin^{-1}\left\{\cos(\tan^{-1} 1)\right\}\right] \)
\( = \cot\left\{\sin^{-1}\left(\cos \frac{\pi}{4}\right)\right\} \)
\( = \cot\left(\sin^{-1} \frac{1}{\sqrt{2}}\right) \)
\( = \cot \frac{\pi}{4} = 1 \)

Question 10. Find the value of the expression \( \sin\left[\cot^{-1}\left\{\cos(\tan^{-1} 1)\right\}\right] \) [3 Marks]
Answer:
\( \sin\left[\cot^{-1}\left\{\cos(\tan^{-1} 1)\right\}\right] \)
\( = \sin\left[\cot^{-1}\left\{\cos \frac{\pi}{4}\right\}\right] \)
\( = \sin\left[\cot^{-1}\left(\frac{1}{\sqrt{2}}\right)\right] \)
\( = \sin\left[\sin^{-1}\left(\frac{\sqrt{2}}{\sqrt{3}}\right)\right] \)
\( = \frac{\sqrt{2}}{\sqrt{3}} \)

Question 11. If \( x = \text{cosec}\left[\tan^{-1}\left\{\cos\left(\cot^{-1}\left(\sec\left(\sin^{-1} a\right)\right)\right)\right\}\right] \) [3 Marks]
and \( y = \sec\left[\cot^{-1}\left\{\sin\left(\tan^{-1}\left(\text{cosec}\left(\cos^{-1} a\right)\right)\right)\right\}\right] \)
where a ∈ [0, 1). Find the relationship between x and y in terms of a.
Answer:
Given \( x = \text{cosec}\left[\tan^{-1}\left\{\cos\left(\cot^{-1}\left(\sec\left(\sin^{-1} a\right)\right)\right)\right\}\right] \)
\( = \text{cosec}\left[\tan^{-1}\left\{\cos\left(\cot^{-1}\left(\sec\left(\sec^{-1} \frac{1}{\sqrt{1 - a^2}}\right)\right)\right)\right\}\right] \)
\( = \text{cosec}\left[\tan^{-1}\left\{\cos\left(\cot^{-1}\left(\frac{1}{\sqrt{1 - a^2}}\right)\right)\right\}\right] \)
\( = \text{cosec}\left[\tan^{-1}\left\{\cos\left(\cos^{-1}\left(\frac{1}{\sqrt{2 - a^2}}\right)\right)\right\}\right] \)
\( = \text{cosec}\left[\tan^{-1}\left(\frac{1}{\sqrt{2 - a^2}}\right)\right] \)
\( = \text{cosec}\left[\text{cosec}^{-1}\left(\sqrt{3 - a^2}\right)\right] \)
\( x = \sqrt{3 - a^2} \)
Also \( y = \sec\left[\cot^{-1}\left\{\sin\left(\tan^{-1}\left(\text{cosec}\left(\cos^{-1} a\right)\right)\right)\right\}\right] \)
\( = \sec\left[\cot^{-1}\left\{\sin\left(\tan^{-1}\left(\text{cosec}\left(\text{cosec}^{-1} \frac{1}{\sqrt{1 - a^2}}\right)\right)\right)\right\}\right] \)
\( = \sec\left[\cot^{-1}\left\{\sin\left(\tan^{-1}\left(\frac{1}{\sqrt{1 - a^2}}\right)\right)\right\}\right] \)
\( = \sec\left[\cot^{-1}\left\{\sin\left(\sin^{-1}\left(\frac{1}{\sqrt{2 - a^2}}\right)\right)\right\}\right] \)
\( = \sec\left[\cot^{-1}\left(\frac{1}{\sqrt{2 - a^2}}\right)\right] \)
\( = \sec\left[\sec^{-1}\left(\sqrt{3 - a^2}\right)\right] \)
\( y = \sqrt{3 - a^2} \)
\( \therefore x^2 = y^2 = 3 - a^2 \)