Question

Solve for x,
${2tan}^{-1}\left(cosx\right)=ta{n}^{-1}\left(2cosecx\right).$

Answer 1

$2ta{n}^{-1}\left(cosx\right)=ta{n}^{-1}\left(2cosecx\right)$

we know that $2ta{n}^{-1}x=ta{n}^{-1}\frac{2x}{1-x^2}$

$\therefore 2ta{n}^{-1}\left(cosx\right)=ta{n}^{-1}\left(2cosecx\right)$

$\Rightarrow ta{n}^{-1}\left(\frac{2cosx}{1-co{s}^{2}x}\right)=ta{n}^{-1}\left(2cosecx\right)$

$\Rightarrow \frac{2cosx}{1-co{s}^{2}x}=2cosecx$

$\Rightarrow \frac{cosx}{si{n}^{2}x}=cosecx$ $[\because si{n}^{2}x+co{s}^{2}x=1]$

$\Rightarrow \frac{cosx}{si{n}^{2}x}=\frac{1}{sinx}$ $[\because cosecx=\frac{1}{sinx}]$

$\Rightarrow \frac{cosx}{sinx}=1$

$\Rightarrow cosx=sinx$

$\Rightarrow cosx=cos(\frac{\pi }{2}-x)$

$\Rightarrow x=\frac{\pi }{2}-x$

$\Rightarrow 2x=\frac{\pi }{2}$

$\Rightarrow x=\frac{\pi }{4}$