Question

Show that $ta{n}^{-1}\frac{1}{3}+ta{n}^{-1}\frac{1}{5}=\frac{1}{2}co{s}^{-1}\frac{33}{65}$

Answer 1

To prove:- ${\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{1}{5}=\frac{1}{2}{\cos }^{-1}\frac{33}{65}$

Proof:-

${\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{1}{5}$

As we know that,

${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$

Therefore,

${\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{1}{5}=\frac{1}{2}{\cos }^{-1}\frac{33}{65}$

${\tan }^{-1}\left(\frac{\frac{1}{3}+\frac{1}{5}}{1-\frac{1}{3}\cdot \frac{1}{5}}\right)=\frac{1}{2}{\cos }^{-1}\frac{33}{65}$

${\tan }^{-1}\left(\frac{\frac{5+3}{15}}{\frac{15-1}{15}}\right)=\frac{1}{2}{\cos }^{-1}\frac{33}{65}$

${\tan }^{-1}\left(\frac{8}{14}\right)=\frac{1}{2}{\cos }^{-1}\frac{33}{65}$

$2{\tan }^{-1}\left(\frac{4}{7}\right)={\cos }^{-1}\frac{33}{65}$

$\Rightarrow {\tan }^{-1}\left(\frac{2\times \frac{4}{7}}{1-{\left(\frac{4}{7}\right)}^2}\right)={\cos }^{-1}\frac{33}{65}$

$\Rightarrow {\tan }^{-1}\left(\frac{\frac{8}{7}}{\frac{49-16}{49}}\right)={\cos }^{-1}\frac{33}{65}$

$\Rightarrow {\tan }^{-1}\frac{56}{33}={\cos }^{-1}\frac{33}{65}$

$\Rightarrow {\cos }^{-1}\frac{33}{65}={\cos }^{-1}\frac{33}{65}(\because {\tan }^{-1}\frac{56}{33}={\cos }^{-1}\frac{33}{65})$

L.H.S. $=$ R.H.S.

Hence proved.