Mark the tick against the correct answer in the following:

To Find: The value of $\cos \left(2 \tan ^{-1} \frac{1}{2}\right)$

Let, $x=\cos \left(2 \tan ^{-1} \frac{1}{2}\right)$

$\Rightarrow x=\cos \left(\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{2}\right)$

Since we know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$

$\Rightarrow \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{2}=\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{2}}{\left.1-\frac{1}{2} \times \frac{1}{2}\right)}\right)=\tan ^{-1} \frac{4}{3}$

$\Rightarrow x=\cos \left(\tan ^{-1} \frac{4}{3}\right)$

Now, let $y=\tan ^{-1} \frac{4}{3}$

$\Rightarrow \tan \mathrm{y}=\frac{4}{3}=\frac{\text { opposite side }}{\text { adjacent side }}$

By pythagorus theroem,

(Hypotenuse ) $^{2}=$ (opposite side $)^{2}+(\text { adjacent side })^{2}$

Therefore, Hypotenuse $=5$

$\Rightarrow \cos \left(\tan ^{-1} \frac{4}{3}\right)=\cos y=\frac{3}{5}$