The value of $\tan \left(\sin ^{-1}\left(\frac{3}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right)$ is
The value of $\tan \left(\sin ^{-1}\left(\frac{3}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right)$ is
- $\frac{6}{17}$
- $\frac{17}{6}$
- $\frac{16}{7}$
- $\frac{7}{16}$
Solution
$\tan \left[\sin ^{-1}\left(\frac{3}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right]$
$=\tan \left[\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right]$
$\cdots\left[\because \sin ^{-1} x=\tan ^{-1} \frac{x}{\sqrt{1-x^2}}\right]$
$=\tan \left[\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \cdot \frac{2}{3}}\right)\right]$
$=\tan \left[\tan ^{-1}\left(\frac{17}{6}\right)\right]$
$=\frac{17}{6}$
Asked in: MHT CET 2023 (13 May Shift 2)
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