cot ^-1(√cos α)=tan ^-1(√cos α)=x, then sin x=

$\cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x$cot−1(cosα​)−tan−1(cosα​)=x $\tan ^{-1}\left(\frac{1}{\sqrt{\cos \alpha}}\right)-\tan ^{-1}(\sqrt{\cos \alpha})=x \Rightarrow \tan ^{-1} \frac{\frac{1}{\sqrt{\cos \alpha}}-\sqrt{\cos \alpha}}{1+\frac{1}{\sqrt{\cos \alpha}} \cdot \sqrt{\cos \alpha}}=x$tan−1(cosα​1​)−tan−1(cosα​)=x⇒tan−11+cosα​1​⋅cosα​cosα​1​−cosα​​=x $\Rightarrow \tan ^{-1} \frac{1-\cos \alpha}{2 \sqrt{\cos \alpha}}=x \Rightarrow \tan x=\frac{1-\cos \alpha}{2 \sqrt{\cos \alpha}}$⇒tan−12cosα​1−cosα​=x⇒tanx=2cosα​1−cosα​ OR $\cot x=\frac{2 \sqrt{\cos \alpha}}{1-\cos \alpha}$cotx=1−cosα2cosα​​ or $\operatorname{cosec} x=\frac{1+\cos \alpha}{1-\cos \alpha}$cosecx=1−cosα1+cosα​ $\sin x=\frac{1-\cos \alpha}{1+\cos \alpha}=\frac{1-\left(1-2 \sin ^2 \alpha / 2\right)}{1+2 \cos ^2 \alpha / 2-1}$sinx=1+cosα1−cosα​=1+2cos2α/2−11−(1−2sin2α/2)​ or $\sin x=\tan ^2 \frac{\alpha}{2}$sinx=tan22α​