Question

$\cos \left(2({\tan }^{-1}\frac{1}{5}+{\tan }^{-1}5)\right)=$ ______.

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

Answer 1

The correct option is D $-1$

We know that ${\tan }^{-1}a+{\tan }^{-1}b={\tan }^{-1}\left(\frac{a+b}{1-ab}\right)$

$\therefore {\tan }^{-1}\frac{1}{5}+{\tan }^{-1}5={\tan }^{-1}\left(\frac{\frac{1}{5}+5}{1-\frac{1}{5}\times 5}\right)={\tan }^{-1}\left(\frac{26}{0}\right)={\tan }^{-1}\infty =\frac{\pi }{2}$

$\therefore \cos \left(2({\tan }^{-1}\frac{1}{5}+{\tan }^{-1}5)\right)=\cos (2\times \frac{\pi }{2})=\cos \pi =-1$
Hence, $\cos \left(2({\tan }^{-1}\frac{1}{5}+{\tan }^{-1}5)\right)=-1$