Question

Prove that $ta{n}^{-1}\left(\frac{\sqrt{1+cosx}+\sqrt{1-cosx}}{\sqrt{1+cosx}-\sqrt{1-cosx}}\right)=\frac{\pi }{4}-\frac{x}{2},\text{where}\pi <x<\frac{3\pi }{2}$

Answer 1

$LHS:ta{n}^{-1}\left(\frac{\sqrt{1+cosx}+\sqrt{1-cosx}}{\sqrt{1+cosx}-\sqrt{1-cosx}}\right)=ta{n}^{-1}\left(\frac{\sqrt{2co{s}^2\frac{x}{2}}+\sqrt{2si{n}^2\frac{x}{2}}}{\sqrt{2co{s}^2\frac{x}{2}-\sqrt{2si{n}^2\frac{x}{2}}}}\right)$

$\Rightarrow =ta{n}^{-1}\left(\frac{\left|cos\frac{x}{2}\right|+\left|sin\frac{x}{2}\right|}{\left|cos\frac{x}{2}\right|-\left|sin\frac{x}{2}\right|}\right)=ta{n}^{-1}\left(\frac{-cos\frac{x}{2}+sin\frac{x}{2}}{-cos\frac{x}{2}-sin\frac{x}{2}}\right)\{\begin{array}{c}\because \pi <x<\frac{3\pi }{2}\Rightarrow \frac{\pi }{2}<\frac{x}{2}<\frac{3\pi }{4}\\ \therefore \frac{x}{2}ϵII\text{quad}\end{array}$
$\Rightarrow =ta{n}^{-1}\left(\frac{1-tan\frac{x}{2}}{1+tan\frac{x}{2}}\right)=ta{n}^{-1}tan(\frac{\pi }{4}-\frac{x}{2})=\frac{\pi }{4}-\frac{x}{2}=\text{RHS.}$