Question

Prove that
${\tan }^{-1}\left(\frac{1}{7}\right)+{\tan }^{-1}\left(\frac{1}{13}\right)={\tan }^{-1}\left(\frac{2}{9}\right)$

Answer 1

Prove thta,

${\tan }^{-1}\left(\frac{1}{7}\right)+{\tan }^{-1}\left(\frac{1}{13}\right)={\tan }^{-1}\left(\frac{2}{9}\right)$

$L.H.S$

${\tan }^{-1}\left(\frac{1}{7}\right)+{\tan }^{-1}\left(\frac{1}{13}\right)$

We know that

${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\frac{x+y}{1-xy}$

using and we get,

${\tan }^{-1}\left(\frac{1}{7}\right)+{\tan }^{-1}\left(\frac{1}{13}\right)={\tan }^{-1}\frac{\frac{1}{7}+\frac{1}{13}}{1-\frac{1}{7}\times \frac{1}{13}}$

$={\tan }^{-1}\frac{\frac{13+7}{91}}{\frac{91-1}{91}}$

$={\tan }^{-1}\frac{20}{90}$

$={\tan }^{-1}\left(\frac{2}{9}\right)$

$R.H.S$

Hence Proved.