Question

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

• A. $\frac{6}{17}$
• B. $\frac{6}{\sqrt{13}}$
• C. $\frac{\sqrt{13}}{5}$
• D. $\frac{17}{6}$

Answer 1

The correct option is D

$\frac{17}{6}$

The explanation for the correct option

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

Let us assume that ${\sin }^{-1}\left(\frac{3}{5}\right)=y$

$$\begin{gathered} \Rightarrow \sin \left(y\right)=\frac{3}{5} \\ \Rightarrow \cos \left(y\right)=\sqrt{1-{\left(\frac{3}{5}\right)}^2}\left[\because {\sin }^2\left(\theta \right)+{\cos }^2\left(\theta \right)=1\right] \\ \Rightarrow \cos \left(y\right)=\sqrt{1-\frac{9}{25}} \\ \Rightarrow \cos \left(y\right)=\sqrt{\frac{25-9}{25}} \\ \Rightarrow \cos \left(y\right)=\sqrt{\frac{16}{25}} \\ \Rightarrow \cos \left(y\right)=\frac{4}{5} \end{gathered}$$

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

$$\begin{gathered} \Rightarrow \tan \left(y\right)=\frac{{\displaystyle \frac{3}{5}}}{{\displaystyle \frac{4}{5}}} \\ \Rightarrow \tan \left(y\right)=\frac{3}{4} \\ \Rightarrow y={\tan }^{-1}\left(\frac{3}{4}\right) \\ \Rightarrow {\sin }^{-1}\left(\frac{3}{5}\right)={\tan }^{-1}\left(\frac{3}{4}\right) \end{gathered}$$

Let us assume that ${\cos }^{-1}\left(\frac{3}{\sqrt{13}}\right)=z$

$$\begin{gathered} \Rightarrow \cos \left(z\right)=\frac{3}{\sqrt{13}} \\ \Rightarrow \sin \left(z\right)=\sqrt{1-{\left(\frac{3}{\sqrt{13}}\right)}^2}\left[\because {\sin }^2\left(\theta \right)+{\cos }^2\left(\theta \right)=1\right] \\ \Rightarrow \sin \left(z\right)=\sqrt{1-\frac{9}{13}} \\ \Rightarrow \sin \left(z\right)=\sqrt{\frac{13-9}{13}} \\ \Rightarrow \sin \left(z\right)=\sqrt{\frac{4}{13}} \\ \Rightarrow \sin \left(z\right)=\frac{2}{\sqrt{13}} \end{gathered}$$

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

$$\begin{gathered} \Rightarrow \tan \left(z\right)=\frac{{\displaystyle \frac{2}{\sqrt{13}}}}{{\displaystyle \frac{3}{\sqrt{13}}}} \\ \Rightarrow \tan \left(z\right)=\frac{2}{3} \\ \Rightarrow z={\tan }^{-1}\left(\frac{2}{3}\right) \\ \Rightarrow {\cos }^{-1}\left(\frac{3}{\sqrt{13}}\right)={\tan }^{-1}\left(\frac{2}{3}\right) \end{gathered}$$

Therefore, $\tan \left[{\sin }^{-1}\left(\frac{3}{5}\right)+{\cos }^{-1}\left(\frac{3}{\sqrt{13}}\right)\right]=\tan \left[{\tan }^{-1}\left(\frac{3}{4}\right)+{\tan }^{-1}\left(\frac{2}{3}\right)\right]$

It is known that ${\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(A\right)\mathbf{+}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(B\right)\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\frac{A+B}{1-AB}\right)$.

$$\begin{gathered} \Rightarrow \tan \left[{\sin }^{-1}\left(\frac{3}{5}\right)+{\cos }^{-1}\left(\frac{3}{\sqrt{13}}\right)\right]=\tan \left[{\tan }^{-1}\left(\frac{{\displaystyle \frac{3}{4}}+{\displaystyle \frac{2}{3}}}{1-\left({\displaystyle \frac{3}{4}}\right)\left({\displaystyle \frac{2}{3}}\right)}\right)\right] \\ \Rightarrow \tan \left[{\sin }^{-1}\left(\frac{3}{5}\right)+{\cos }^{-1}\left(\frac{3}{\sqrt{13}}\right)\right]=\frac{{\displaystyle \frac{\left(3\times 3\right)+\left(2\times 4\right)}{12}}}{1-{\displaystyle \frac{1}{2}}} \\ \Rightarrow \tan \left[{\sin }^{-1}\left(\frac{3}{5}\right)+{\cos }^{-1}\left(\frac{3}{\sqrt{13}}\right)\right]=\frac{{\displaystyle \frac{9+8}{12}}}{{\displaystyle \frac{1}{2}}} \\ \Rightarrow \tan \left[{\sin }^{-1}\left(\frac{3}{5}\right)+{\cos }^{-1}\left(\frac{3}{\sqrt{13}}\right)\right]=\frac{17}{12}\times 2 \\ \Rightarrow \tan \left[{\sin }^{-1}\left(\frac{3}{5}\right)+{\cos }^{-1}\left(\frac{3}{\sqrt{13}}\right)\right]=\frac{17}{6} \end{gathered}$$

Hence, option D is correct.