The value of tan left(2 tan ^{-1}left(frac{3} | vedclass

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(D) Let $\alpha = 	an^{-1}\left(\frac{3}{5}ight)$ and $\beta = \sin^{-1}\left(\frac{5}{13}ight)$.We need to find $	an(2\alpha + \beta)$.First,co

Vedclass · Accepted Answer

$\frac{220}{21}$

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(D) Let $\alpha = 	an^{-1}\left(\frac{3}{5}ight)$ and $\beta = \sin^{-1}\left(\frac{5}{13}ight)$.We need to find $	an(2\alpha + \beta)$.First,convert $\beta$ to $	an^{-1}$ form: Since $\sin \beta = \frac{5}{13}$,then $	an \beta = \frac{5}{\sqrt{13^2 - 5^2}} = \frac{5}{12}$. So,$\beta = 	an^{-1}\left(\frac{5}{12}ight)$.Now,$2\alpha = 2	an^{-1}\left(\frac{3}{5}ight) = 	an^{-1}\left(\frac{2 \cdot \frac{3}{5}}{1 - (\frac{3}{5})^2}ight) = 	an^{-1}\left(\frac{6/5}{1 - 9/25}ight) = 	an^{-1}\left(\frac{6/5}{16/25}ight) = 	an^{-1}\left(\frac{6}{5} \cdot \frac{25}{16}ight) = 	an^{-1}\left(\frac{15}{8}ight)$.Now we need $	an(2\alpha + \beta) = 	an\left(	an^{-1}\left(\frac{15}{8}ight) + 	an^{-1}\left(\frac{5}{12}ight)ight)$.Using the formula $	an^{-1} x + 	an^{-1} y = 	an^{-1}\left(\frac{x+y}{1-xy}ight)$:$= 	an\left(	an^{-1}\left(\frac{\frac{15}{8} + \frac{5}{12}}{1 - \frac{15}{8} \cdot \frac{5}{12}}ight)ight) = \frac{\frac{45+10}{24}}{1 - \frac{75}{96}} = \frac{55/24}{21/96} = \frac{55}{24} \cdot \frac{96}{21} = \frac{55 \cdot 4}{21} = \frac{220}{21}$.

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

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