Find the value of $\sin \left( 2 \tan^{-1} \frac{1}{3} \right) + \cos(\tan^{-1} 2\sqrt{2})$.

$\frac{14}{15}$

The correct answer is Option (1) → $\frac{14}{15}$ ##

We have, $\sin \left( 2 \tan^{-1} \frac{1}{3} \right) + \cos(\tan^{-1} 2\sqrt{2})$

Let $\tan^{-1} \frac{1}{3} = x \Rightarrow \tan x = \frac{1}{3} \dots \text{(i)}$

Again, let $\tan^{-1} 2\sqrt{2} = y \Rightarrow \tan y = 2\sqrt{2} \dots \text{(ii)}$

$∴\sin \left( 2 \tan^{-1} \frac{1}{3} \right) + \cos(\tan^{-1} 2\sqrt{2}) = \sin(2x) + \cos(y)$

$= \frac{2 \tan x}{1 + \tan^2 x} + \frac{1}{\sec y} \quad \left[ ∵\sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} \right]$

$= \frac{2 \times \frac{1}{3}}{1 + \left( \frac{1}{3} \right)^2} + \frac{1}{\sqrt{1 + \tan^2 y}} \quad [\text{from Eq. (i) and } 1 + \tan^2 \theta = \sec^2 \theta]$

$= \frac{2/3}{10/9} + \frac{1}{\sqrt{1 + (2\sqrt{2})^2}} \quad [\text{from Eq. (ii)}]$

$= \frac{3}{5} + \frac{1}{\sqrt{9}} = \frac{3}{5} + \frac{1}{3} = \frac{9 + 5}{15} = \frac{14}{15}$