Question

$\cos \left(2\tan {}^{-1}\frac{1}{7}\right)-\sin \left(4\tan {}^{-1}\frac{1}{3}\right)=$

• A. $1$
• B. $\frac{1}{2}$
• C. $-\frac{1}{2}$
• D. $0$

Answer 1

The correct option is D $0$

$\cos \left(2{\tan }^{-1}\frac{1}{7}\right)-\sin \left(4{\tan }^{-1}\frac{1}{3}\right)$
Take $2{\tan }^{-1}\frac{1}{7}=y$ then $\tan y/2=\frac{1}{7}$
hence $\sin y/2=\frac{1}{\sqrt{50}}$ and $\cos y/2=\frac{7}{\sqrt{50}}$
Therefore , $siny=2\left(cosy/2\right)\left(siny/2\right)=\frac{7}{\sqrt{50}}\times \frac{1}{\sqrt{50}}=\frac{7}{25}$
$=\cos \left({\sin }^{-1}\frac{7}{25}\right)-\sin \left({\sin }^{-1}\frac{24}{25}\right)$
$=\cos \left({\cos }^{-1}\frac{24}{25}\right)-\sin \left({\sin }^{-1}\frac{24}{25}\right)$
$=\frac{24}{25}-\frac{24}{25}=0$