JEE PYQ: Inverse Trigonometric Functions Question 12
Question 12 - 2020 (03 Sep Shift 1)
$2\pi - \left(\sin^{-1}\frac{4}{5} + \sin^{-1}\frac{5}{13} + \sin^{-1}\frac{16}{65}\right)$ is equal to:
(1) $\frac{\pi}{2}$
(2) $\frac{5\pi}{4}$
(3) $\frac{3\pi}{2}$
(4) $\frac{7\pi}{4}$
Show Answer
Answer: (3)
Solution
$\sin^{-1}\frac{4}{5} = \tan^{-1}\frac{4}{3}$, $\sin^{-1}\frac{5}{13} = \tan^{-1}\frac{5}{12}$. Then $\tan^{-1}\frac{4}{3} + \tan^{-1}\frac{5}{12} = \tan^{-1}\left(\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3}\cdot\frac{5}{12}}\right) = \tan^{-1}\frac{63}{16}$. And $\tan^{-1}\frac{63}{16} + \cot^{-1}\frac{63}{16} = \frac{\pi}{2}$, with $\sin^{-1}\frac{16}{65} = \tan^{-1}\frac{16}{63}$. So the sum $= \frac{\pi}{2}$. Answer: $2\pi - \frac{\pi}{2} = \frac{3\pi}{2}$.
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