The value of $\tan \left(2 \tan ^{-1}\left(\frac{3}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)\right)$ is equal to :

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Inverse Trigonometric Functions

The value of

tan (2 tan^{- 1} (\frac{3}{5}) + sin^{- 1} (\frac{5}{13}))

is equal to :

Answer: (C)

With

A = tan^{- 1} \frac{3}{5}

tan 2 A = \frac{2 \cdot \frac{3}{5}}{1 - \frac{9}{25}} = \frac{\frac{6}{5}}{\frac{16}{25}} = \frac{15}{8}

and with

B = sin^{- 1} \frac{5}{13}

, the right triangle with legs

5

and

12

gives

tan B = \frac{5}{12}

. Therefore

tan (2 A + B) = \frac{\frac{15}{8} + \frac{5}{12}}{1 - \frac{15}{8} \cdot \frac{5}{12}} = \frac{\frac{55}{24}}{\frac{21}{96}} = \frac{220}{21}

so the value is

\frac{220}{21}

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