CBSE Class 12 Mathematics Inverse Trigonometric Functions Previous Year Questions Set 01

This page has CBSE board Previous Year Questions with complete step-wise solutions for Class 12 Mathematics Chapter 2, Inverse Trigonometric Functions. The questions are grouped by marks and focus on principal values and evaluating inverse trigonometric expressions.

Very Short Answer Questions (1 Mark)

Question 1. Using principal value, evaluate the following: \( \cos^{-1}\left(\cos \frac{2\pi}{3}\right) + \sin^{-1}\left(\sin \frac{2\pi}{3}\right) \) [CBSE 2008, 11, 1M]
Answer:
\( \cos^{-1}\left(\cos \frac{2\pi}{3}\right) + \sin^{-1}\left(\sin \frac{2\pi}{3}\right) \)
\( = \cos^{-1}\left(\cos \frac{2\pi}{3}\right) + \sin^{-1}\left[\sin\left(\pi - \frac{\pi}{3}\right)\right] \)
\( = \frac{2\pi}{3} + \frac{\pi}{3} = \pi \) [1]

Question 2. Write the principal value of \( \cos^{-1}\left(\cos \frac{7\pi}{6}\right) \) [CBSE 2009, 1M]
Answer:
\( \cos^{-1}\left[\cos\left(2\pi - \frac{5\pi}{6}\right)\right] = \frac{5\pi}{6} \) [1]

Question 3. Write the principal value of \( \sec^{-1}(-2) \). [CBSE 2010, 1M]
Answer:
Let \( \sec^{-1}(-2) = \theta \Rightarrow \sec\theta = -2 \)
\( \Rightarrow \cos\theta = -\frac{1}{2} = \cos \frac{2\pi}{3} \)
\( \Rightarrow \theta = \frac{2\pi}{3} \)
\( \Rightarrow \sec^{-1}(-2) = \frac{2\pi}{3} \) [1]

Question 4. Write the principal value of \( \cot^{-1}(-\sqrt{3}) \). [CBSE 2010, 1M]
Answer:
Let \( \cot^{-1}(-\sqrt{3}) = \theta \Rightarrow \cot\theta = -\sqrt{3} \)
\( \Rightarrow \cot\theta = \cot\left(\pi - \frac{\pi}{6}\right) = \cot \frac{5\pi}{6} \)
\( \Rightarrow \theta = \frac{5\pi}{6} \Rightarrow \cot^{-1}(-\sqrt{3}) = \frac{5\pi}{6} \) [1]

Question 5. Find the value of \( \sin^{-1}\left(\sin \frac{4\pi}{5}\right) \). [CBSE 2010, 1M]
Answer:
\( \sin^{-1}\left(\sin \frac{4\pi}{5}\right) \)
\( = \sin^{-1}\left[\sin\left(\pi - \frac{\pi}{5}\right)\right] \)
\( = \sin^{-1}\left(\sin \frac{\pi}{5}\right) = \frac{\pi}{5} \) [1]

Question 6. Write the value of \( \tan^{-1}\left[2\sin\left(2\cos^{-1} \frac{\sqrt{3}}{2}\right)\right] \). [CBSE 2013, 1M]
Answer:
\( \tan^{-1}\left[2\sin\left(2\cos^{-1} \frac{\sqrt{3}}{2}\right)\right] \)
\( = \tan^{-1}\left[2\sin\left(2 \times \frac{\pi}{6}\right)\right] \)
\( = \tan^{-1}\left[2\sin \frac{\pi}{3}\right] \)
\( = \tan^{-1}\left[2 \times \frac{\sqrt{3}}{2}\right] = \tan^{-1} \sqrt{3} = \frac{\pi}{3} \) [1]

Question 7. Find the value of \( \tan^{-1} \sqrt{3} - \cot^{-1}(-\sqrt{3}) \). [CBSE 2013, 18, 1M]
Answer:
Let \( \cot^{-1}(-\sqrt{3}) = \theta \Rightarrow \cot\theta = -\sqrt{3} \)
\( \Rightarrow \cot\theta = \cot\left(\pi - \frac{\pi}{6}\right) \Rightarrow \theta = \frac{5\pi}{6} \)
\( \therefore \tan^{-1} \sqrt{3} - \cot^{-1}(-\sqrt{3}) \)
\( = \frac{\pi}{3} - \frac{5\pi}{6} \)
\( = \frac{\pi}{3} - \pi + \frac{\pi}{6} \)
\( = -\frac{\pi}{2} \) [1]

Question 8. The range of the principal value branch of the function \( y = \sec^{-1} x \) is ............... . [CBSE 2020, 1M]
OR
The principal value of \( \cos^{-1}\left(-\frac{1}{2}\right) \) is ............... .
Answer:
\( y = \sec^{-1} x \)
Range of \( \sec^{-1} x \) is \( [0, \pi] - \left\{\frac{\pi}{2}\right\} \) [1]
OR
\( \cos^{-1}\left(-\frac{1}{2}\right) = \theta \Rightarrow \cos\theta = -\frac{1}{2} = \cos\left(\pi - \frac{\pi}{3}\right) = \cos \frac{2\pi}{3} \)
\( \Rightarrow \theta = \frac{2\pi}{3} \Rightarrow \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3} \) [1]

Long Answer-I Questions (4 Marks)

Question 9. Evaluate: \( \tan\left\{2\tan^{-1}\left(\frac{1}{5}\right) + \frac{\pi}{4}\right\} \) [CBSE 2015, 4M]
Answer:
Let \( \tan^{-1} \frac{1}{5} = \theta \Rightarrow \tan\theta = \frac{1}{5} \)
\( \therefore \tan\left\{2\tan^{-1}\left(\frac{1}{5}\right) + \frac{\pi}{4}\right\} \)
\( = \tan\left[2\theta + \frac{\pi}{4}\right] \) [1]
\( = \frac{1 + \tan 2\theta}{1 - \tan 2\theta} = \frac{1 + \left(\frac{2\tan\theta}{1 - \tan^2\theta}\right)}{1 - \left(\frac{2\tan\theta}{1 - \tan^2\theta}\right)} \) [1]
\( = \frac{1 + \left(\frac{2/5}{1 - 1/25}\right)}{1 - \left(\frac{2/5}{1 - 1/25}\right)} \) [1]
\( = \frac{1 + \frac{5}{12}}{1 - \frac{5}{12}} = \frac{17}{7} \) [1]