Question

Prove that ${\cot }^{-1}9+{\text{cosec}}^{-1}\frac{\sqrt{41}}{4}=\frac{\pi }{4}$

Answer 1

$x={\cot }^{-1}9$

$\Rightarrow \cot x=9$

$\Rightarrow \tan x=\frac{1}{9}$

$x={\tan }^{-1}\frac{1}{9}$

$y={cosec}^{-1}\frac{\sqrt{41}}{4}$

$\Rightarrow cosecx=\frac{\sqrt{41}}{4}$

$\tan y=\frac{4}{5}$

$y={\tan }^{-1}\frac{4}{5}$

$Lhs={\tan }^{-1}\frac{1}{9}+\tan \frac{4}{5}$

$-{\tan }^{-1}\frac{\frac{1}{9}+\frac{4}{5}}{1-\frac{4}{45}}={\tan }^{-1}\frac{41}{41}$

$={\tan }^{-1}1=\frac{\pi }{4}=RHS$