Question

Find the value of: $\sin \left(2{\tan }^{-1}\frac{1}{4}\right)+\cos \left({\tan }^{-1}2\sqrt{2}\right)$

Answer 1

Given $\sin \left(2{\tan }^{-1}\frac{1}{4}\right)+\cos \left({\tan }^{-1}2\sqrt{2}\right)----\left(1\right)$

$\sin \left(2{\tan }^{-1}\frac{1}{4}\right)=\sin \left({\sin }^{-1}\frac{2\left(\frac{1}{4}\right)}{\left(\frac{1}{4}\right)}\right)[\because 2{\tan }^{-1}x={\sin }^{-1}\frac{2x}{1+x^2}]$

$\sin \left({\sin }^{-1}\frac{2/4}{17/16}\right)$

$=\sin \left({\sin }^{-1}\frac{8}{17}\right)$

$=\frac{8}{17}$

$\cos \left({\tan }^{-1}2\sqrt{2}\right)$

Let ${\tan }^{-1}2\sqrt{2}=\lambda$

$\tan \lambda =2\sqrt{2}$

$\cos \lambda =\frac{1}{\sqrt{1+{\tan }^2\lambda }}=\frac{1}{\sqrt{1+{\left(2\sqrt{2}\right)}^2}}=\frac{1}{\sqrt{1+8}}=\frac{1}{\sqrt{9}}$

$\cos \lambda =\frac{1}{3}$ i.e; $\cos \left({\tan }^{-1}2\sqrt{2}\right)=\frac{1}{3}$

From $\left(1\right)\sin \left(2{\tan }^{-1}\frac{1}{4}\right)+\cos \left({\tan }^{-1}2\sqrt{2}\right)$

$=\frac{8}{17}+\frac{1}{3}=\frac{24+17}{51}=\frac{41}{51}$