Question

Prove that ${\cos }^{-1}\left(\frac{4}{5}\right)+{\tan }^{-1}\left(\frac{3}{5}\right)={\tan }^{-1}\left(\frac{27}{11}\right)$

Answer 1

L.H.S

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

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

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

Therefore,

$={\tan }^{-1}\left(\frac{\frac{3}{4}+\frac{3}{5}}{1-\frac{3}{4}\times \frac{3}{5}}\right)$

$={\tan }^{-1}\left(\frac{15+12}{20-9}\right)$

$={\tan }^{-1}\left(\frac{27}{11}\right)$

$=$R.H.S

Hence, proved.