JEE PYQ: Inverse Trigonometric Functions Question 14
Question 14 - 2019 (08 Apr Shift 1)
If $\alpha = \cos^{-1}\left(\frac{3}{5}\right)$, $\beta = \tan^{-1}\left(\frac{1}{3}\right)$, where $0 < \alpha, \beta < \frac{\pi}{2}$, then $\alpha - \beta$ is equal to:
(1) $\tan^{-1}\left(\frac{9}{5\sqrt{10}}\right)$
(2) $\cos^{-1}\left(\frac{9}{5\sqrt{10}}\right)$
(3) $\tan^{-1}\left(\frac{9}{14}\right)$
(4) $\sin^{-1}\left(\frac{9}{5\sqrt{10}}\right)$
Show Answer
Answer: (4)
Solution
$\cos\alpha = \frac{3}{5}$, so $\tan\alpha = \frac{4}{3}$. $\tan\beta = \frac{1}{3}$. $\tan(\alpha - \beta) = \frac{\frac{4}{3} - \frac{1}{3}}{1 + \frac{4}{3}\cdot\frac{1}{3}} = \frac{1}{\frac{13}{9}} = \frac{9}{13}$. So $\alpha - \beta = \tan^{-1}\left(\frac{9}{13}\right) = \sin^{-1}\left(\frac{9}{5\sqrt{10}}\right) = \cos^{-1}\left(\frac{13}{5\sqrt{10}}\right)$.
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