Question

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

• A. $\frac{220}{21}$
• B. $\frac{110}{21}$
• C. $\frac{55}{21}$
• D. $\frac{20}{11}$

Answer 1

The correct option is A

$\frac{220}{21}$

The explanation for the correct option

The given trigonometric expression: $\tan \left[2{\tan }^{-1}\left(\frac{3}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)\right]$.

It is known that, $2{\tan }^{-1}\left(x\right)={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$.

Thus, $2{\tan }^{-1}\left(\frac{3}{5}\right)={\tan }^{-1}\left(\frac{2\times {\displaystyle \frac{3}{5}}}{1-{\left({\displaystyle \frac{3}{5}}\right)}^2}\right)$.

$$\begin{gathered} \Rightarrow 2{\tan }^{-1}\left(\frac{3}{5}\right)={\tan }^{-1}\left(\frac{{\displaystyle \frac{6}{5}}}{1-{\displaystyle \frac{9}{25}}}\right) \\ \Rightarrow 2{\tan }^{-1}\left(\frac{3}{5}\right)={\tan }^{-1}\left(\frac{{\displaystyle \frac{6}{5}}}{{\displaystyle \frac{16}{25}}}\right) \\ \Rightarrow 2{\tan }^{-1}\left(\frac{3}{5}\right)={\tan }^{-1}\left(\frac{6}{5}\times \frac{25}{16}\right) \\ \Rightarrow 2{\tan }^{-1}\left(\frac{3}{5}\right)={\tan }^{-1}\left(\frac{15}{8}\right) \end{gathered}$$

Let us assume that, ${\sin }^{-1}\left(\frac{5}{13}\right)=y$

$$\begin{gathered} \Rightarrow \sin \left(y\right)=\frac{5}{13} \\ \Rightarrow {\sin }^2\left(y\right)={\left(\frac{5}{13}\right)}^2 \\ \Rightarrow 1-{\cos }^2\left(y\right)=\frac{25}{169}\left[\because {\sin }^2\left(\theta \right)+{\cos }^2\left(\theta \right)=1\right] \\ \Rightarrow {\cos }^2\left(y\right)=1-\frac{25}{169} \\ \Rightarrow \cos \left(y\right)=\frac{169-25}{169} \\ \Rightarrow \cos \left(y\right)=\sqrt{\frac{144}{169}} \\ \Rightarrow \cos \left(y\right)=\frac{12}{13} \end{gathered}$$

Thus, $\tan \left(y\right)=\frac{\sin \left(y\right)}{\cos \left(y\right)}$.

$$\begin{gathered} \Rightarrow \tan \left(y\right)=\frac{{\displaystyle \frac{5}{13}}}{{\displaystyle \frac{12}{13}}} \\ \Rightarrow \tan \left(y\right)=\frac{5}{12} \\ \Rightarrow y={\tan }^{-1}\left(\frac{5}{12}\right) \\ \Rightarrow {\sin }^{-1}\left(\frac{5}{13}\right)={\tan }^{-1}\left(\frac{5}{12}\right) \end{gathered}$$

Therefore, $\tan \left[2{\tan }^{-1}\left(\frac{3}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)\right]=\tan \left[{\tan }^{-1}\left(\frac{15}{8}\right)+{\tan }^{-1}\left(\frac{5}{12}\right)\right]$.

$$\begin{gathered} \Rightarrow \tan \left[2{\tan }^{-1}\left(\frac{3}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)\right]=\tan \left[{\tan }^{-1}\left(\frac{\frac{15}{8}+\frac{5}{12}}{1-\left(\frac{15}{8}\right)\left(\frac{5}{12}\right)}\right)\right]\left[\because {\tan }^{-1}\left(A\right)+{\tan }^{-1}\left(B\right)={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)\right] \\ \Rightarrow \tan \left[2{\tan }^{-1}\left(\frac{3}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)\right]=\left(\frac{\frac{45+10}{24}}{1-{\displaystyle \frac{25}{32}}}\right) \\ \Rightarrow \tan \left[2{\tan }^{-1}\left(\frac{3}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)\right]=\left(\frac{\frac{55}{24}}{\frac{7}{32}}\right) \\ \Rightarrow \tan \left[2{\tan }^{-1}\left(\frac{3}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)\right]=\left(\frac{55}{24}\times \frac{32}{7}\right) \\ \Rightarrow \tan \left[2{\tan }^{-1}\left(\frac{3}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)\right]=\frac{220}{21} \end{gathered}$$

Hence. option A is correct.