The value of $\tan \left(\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right)$ is
The value of $\tan \left(\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right)$ is
- $\frac{6}{17}$
- $\frac{7}{16}$
- $\frac{16}{7}$
- $\frac{17}{6}$
Solution
$\tan \left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)=\tan \left(\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{2}{3}\right)$
$=\tan \tan ^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \times \frac{2}{3}}\right)=\tan \tan ^{-1}\left(\frac{17}{6}\right)=\frac{17}{6}$
Asked in: MHT CET 2022 (10 Aug Shift 2)
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