Question

$Showthat$ $co{t}^{-1}\left(\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}\right)=\frac{x}{2}$ $forx\in (0,\frac{\pi }{2})$

Answer 1

$\sqrt{1+sinx}=\sqrt{si{n}^2\left(\frac{x}{2}\right)+co{s}^2\left(\frac{x}{2}\right)+2si{n}^2\left(\frac{x}{2}\right)co{s}^2\left(\frac{x}{2}\right)}=\sqrt{\left(sin\right(\frac{x}{2})+cos(\frac{x}{2}){)}^2}=\left(sin\right(\frac{x}{2})+cos(\frac{x}{2})\sqrt{1-sinx}=\sqrt{si{n}^2\left(\frac{x}{2}\right)+co{s}^2\left(\frac{x}{2}\right)-2si{n}^2\left(\frac{x}{2}\right)co{s}^2\left(\frac{x}{2}\right)}=\sqrt{\left(cos\right(\frac{x}{2})-sin(\frac{x}{2}){)}^2}=(cos\left(\frac{x}{2}\right)-sin\left(\frac{x}{2}\right))\therefore co{t}^{-1}(\left(\frac{\left(cos\right(\frac{x}{2})+sin(\frac{x}{2})+cos(\frac{x}{2})-sin(\frac{x}{2})}{\left(cos\right(\frac{x}{2})+sin(\frac{x}{2})-(cos\left(\frac{x}{2}\right)+sin\left(\frac{x}{2}\right)}\right))=co{t}^{-1}(cot\left(\frac{x}{2}\right))butthisvalidonlyforx\in (0,\left(\frac{\pi }{2}\right))$