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Inverse Trigonometric Functions EXERCISE 2.MISC Solutions
- Q1: Find the value of the following: $\cos^{-1}(\cos(\frac{13\pi}{6}))$
- Q2: Find the value of the following: $\tan^{-1}(\tan(\frac{7\pi}{6}))$
- Q3: Prove that $2\sin^{-1} \frac{3}{5} = \tan^{-1} \frac{24}{7}$
- Q4: Prove that $\sin^{-1} \frac{8}{17} + \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{77}{36}$
- Q5: Prove that $\cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13} = \cos^{-1} \frac{33}{65}$
- Q6: Prove that $\cos^{-1} \frac{12}{13} + \sin^{-1} \frac{3}{5} = \sin^{-1} \frac{56}{65}$
- Q7: Prove that $\tan^{-1} \frac{63}{16} = \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5}$
- Q8: Prove that $\tan^{-1} \sqrt{x} = \frac{1}{2} \cos^{-1} (\frac{1-x}{1+x})$, $x \in [0, 1]$
- Q9: Prove that $\cot^{-1} (\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}) = \frac{x}{2}$, $x \in (0, \frac{\pi}{4})$
- Q10: Prove that $\tan^{-1}(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}) = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x$, $\frac{-1}{\sqrt{2}} \leq x \leq 1$
- Q11: Solve the following equations: $2\tan^{-1}(\cos x) = \tan^{-1}(2 \csc x)$
- Q12: Solve the following equations: $\tan^{-1} \frac{1-x}{1+x} = \frac{1}{2} \tan^{-1} x, (x > 0)$
- Q13: $\sin(\tan^{-1} x)$, $|x| < 1$ is equal to
- Q14: $\sin^{-1}(1 - x) - 2\sin^{-1} x = \frac{\pi}{2}$, then $x$ is equal to
Other Exercises for CBSE - Class 12 Mathematics Inverse Trigonometric Functions
Chapters in CBSE - Class 12 Mathematics
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