Question

Write the following in the simplest form,

$ta{n}^{-1}\left(\sqrt{\frac{1-cosx}{1+cosx}}\right),x<\pi$

Answer 1

$ta{n}^{-1}\left(\sqrt{\frac{1-cosx}{1+cosx}}\right)=ta{n}^{-1}\left(\sqrt{\frac{2si{n}^2\left(\frac{x}{2}\right)}{2co{s}^2\left(\frac{x}{2}\right)}}\right)$
$\left(\begin{array}{c}\because 1-cosx=2si{n}^2\left(\frac{x}{2}\right)\\ and1+cosx=2co{s}^2\left(\frac{x}{2}\right)\end{array}\right)$
$=ta{n}^{-1}\left(\sqrt{ta{n}^2}\frac{x}{2}\right)=ta{n}^{-1}\left(tan\frac{x}{2}\right)=\frac{x}{2}$

Note: If x is positive, then $\sqrt{ta{n}^2\frac{x}{2}}=tan\frac{x}{2}$ and if x is negative, then $\sqrt{ta{n}^2\frac{x}{2}}=-tan\frac{x}{2}$.