Question

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

• A. $\frac{16}{15}$
• B. $\frac{14}{15}$
• C. $\frac{12}{15}$
• D. $\frac{11}{15}$

Answer 1

The correct option is B

$\frac{14}{15}$

Explanation for the correct option

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

Let us assume that, $2{\tan }^{-1}\left(\frac{1}{3}\right)=x$

$$\begin{gathered} \Rightarrow {\tan }^{-1}\left(\frac{1}{3}\right)=\frac{x}{2} \\ \Rightarrow \tan \left(\frac{x}{2}\right)=\frac{1}{3} \end{gathered}$$

It is known that, $\mathbf{sin}\mathbf{\left(}\mathbf{2}\mathbf{\theta }\mathbf{\right)}\mathbf{=}\frac{\mathbf{2}\mathbf{tan}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}}{\mathbf{1}\mathbf{+}{\mathbf{tan}}^{\mathbf{2}}\mathbf{\left(}{\displaystyle \theta }\mathbf{\right)}}$.

Thus, $\sin \left(x\right)=\frac{2\tan \left({\displaystyle \frac{x}{2}}\right)}{1+{\tan }^2\left({\displaystyle \frac{x}{2}}\right)}$.

$$\begin{gathered} \Rightarrow \sin \left(x\right)=\frac{2\left({\displaystyle \frac{1}{3}}\right)}{1+{\left({\displaystyle \frac{1}{3}}\right)}^2}\left[\because \tan \left(\frac{x}{2}\right)=\frac{1}{3}\right] \\ \Rightarrow \sin \left(x\right)=\frac{{\displaystyle \frac{2}{3}}}{1+{\displaystyle \frac{1}{9}}} \\ \Rightarrow \sin \left(x\right)=\frac{{\displaystyle \frac{2}{3}}}{{\displaystyle \frac{9+1}{9}}} \\ \Rightarrow \sin \left(x\right)=\frac{{\displaystyle \frac{2}{3}}}{{\displaystyle \frac{10}{9}}} \\ \Rightarrow \sin \left(x\right)=\frac{2}{3}\times \frac{9}{10} \\ \Rightarrow \sin \left(x\right)=\frac{{\displaystyle 3}}{{\displaystyle 5}} \\ \Rightarrow x={\sin }^{-1}\left(\frac{3}{5}\right) \\ \Rightarrow 2{\tan }^{-1}\left(\frac{1}{3}\right)={\sin }^{-1}\left(\frac{3}{5}\right)\left[\because 2{\tan }^{-1}\left(\frac{1}{3}\right)=x\right] \end{gathered}$$

Let us assume that, ${\tan }^{-1}\left(2\sqrt{2}\right)=y$

$$\begin{gathered} \Rightarrow \tan \left(y\right)=2\sqrt{2} \\ \Rightarrow {\tan }^2\left(y\right)={\left(2\sqrt{2}\right)}^2 \\ \Rightarrow {\sec }^2\left(y\right)-1=8\left[\because {\sec }^2\left(\theta \right)-{\tan }^2\left(\theta \right)=1\right] \\ \Rightarrow {\sec }^2\left(y\right)=9 \\ \Rightarrow \sec \left(y\right)=\sqrt{9} \\ \Rightarrow \sec \left(y\right)=3 \\ \Rightarrow \frac{1}{\sec \left(y\right)}=\frac{1}{3} \\ \Rightarrow \cos \left(y\right)=\frac{1}{3}\left[\because \sec \left(\theta \right)=\frac{1}{\cos \left(\theta \right)}\right] \\ \Rightarrow y={\cos }^{-1}\left(\frac{1}{3}\right) \\ \Rightarrow {\tan }^{-1}\left(2\sqrt{2}\right)={\cos }^{-1}\left(\frac{1}{3}\right)\left[\because {\tan }^{-1}\left(2\sqrt{2}\right)=y\right] \end{gathered}$$

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

$$\begin{gathered} \Rightarrow \sin \left[2{\tan }^{-1}\left(\frac{1}{3}\right)\right]+\cos \left[{\tan }^{-1}\left(2\sqrt{2}\right)\right]=\frac{3}{5}+\frac{1}{3} \\ \Rightarrow \sin \left[2{\tan }^{-1}\left(\frac{1}{3}\right)\right]+\cos \left[{\tan }^{-1}\left(2\sqrt{2}\right)\right]=\frac{\left(3\times 3\right)+\left(1\times 5\right)}{15} \\ \Rightarrow \sin \left[2{\tan }^{-1}\left(\frac{1}{3}\right)\right]+\cos \left[{\tan }^{-1}\left(2\sqrt{2}\right)\right]=\frac{9+5}{15} \\ \Rightarrow \sin \left[2{\tan }^{-1}\left(\frac{1}{3}\right)\right]+\cos \left[{\tan }^{-1}\left(2\sqrt{2}\right)\right]=\frac{14}{15} \end{gathered}$$

Therefore, the value of $\sin \left[2{\tan }^{-1}\left(\frac{1}{3}\right)\right]+\cos \left[{\tan }^{-1}\left(2\sqrt{2}\right)\right]$ is $\frac{14}{15}$.

Hence, the correct option is (B).