Question

Prove that ${\cot }^{-1}9+cose{c}^{-1}\frac{41}{4}=\frac{\pi }{4}$

Answer 1

$LHS$

$co{t}^{-1}9=ta{n}^{-1}\frac{1}{9}$ and $cose{c}^{-1}\frac{\sqrt{4}1}{4}=ta{n}^{-1}\frac{4}{5}$

$co{t}^{-1}9+cose{c}^{-1}\frac{\sqrt{4}1}{4}=ta{n}^{-1}\frac{1}{9}+ta{n}^{-1}\frac{4}{5}=ta{n}^{-1}\left(\frac{\frac{1}{9}+\frac{4}{5}}{1-\frac{1}{9}\times \frac{4}{5}}\right)$

$=ta{n}^{-1}\left(1\right)=\frac{\pi }{4}=RHS$

$\text{hence proved}$