Question

Solve the following equation for x:
$2ta{n}^{-1}\left(cosx\right)=ta{n}^{-1}\left(2cosecx\right)$

Answer 1

Given $2ta{n}^{-1}\left(cosx\right)=ta{n}^{-1}\left(2cosecx\right)\Rightarrow ta{n}^{-1}\left(\frac{2cosx}{1-co{s}^{2}x}\right)=ta{n}^{-1}\left(2cosecx\right)[\because 2ta{n}^{-1}x=ta{n}^{-1}\left(\frac{2x}{1-x^2}\right)]\Rightarrow ta{n}^{-1}\left(\frac{2cosx}{si{n}^{2}x}\right)=ta{n}^{-1}\left(\frac{2}{sinx}\right)\Rightarrow \frac{2cosx}{si{n}^{2}x}=\frac{2}{sinx}\Rightarrow \frac{2cosx}{sinx}=2\Rightarrow cotx=1\Rightarrow cotx=cot\frac{\pi }{4}\Rightarrow x=\frac{\pi }{4}$