Question

If $y={\cot }^{-1}\left\{\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right\}$, show that $\frac{dy}{dx}$ is independent of x.

Answer 1

$$\begin{gathered} \mathrm{Let},y=co{t}^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]...\left(\mathrm{i}\right) \\ \mathrm{We}\mathrm{have},\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \\ =\frac{{\left(\sqrt{1+\sin x}+\sqrt{1-\sin x}\right)}^2}{\left(\sqrt{1+\sin x}-\sqrt{1-\sin x}\right)\left(\sqrt{1+\sin x}+\sqrt{1-\sin x}\right)} \\ =\frac{\left(1+\sin x\right)+\left(1-\sin x\right)+2\sqrt{\left(1-\sin x\right)\left(1+\sin x\right)}}{\left(1+\sin x\right)-\left(1-\sin x\right)} \\ =\frac{2+2\sqrt{1-{\sin }^{2}x}}{2\sin x} \\ =\frac{1+\cos x}{\sin x} \\ =\frac{2{\cos }^2{\displaystyle \frac{x}{2}}}{2\sin {\displaystyle \frac{x}{2}}\cos {\displaystyle \frac{x}{2}}} \\ =cot\frac{x}{2} \\ \mathrm{Therefore},\mathrm{equation}\left(\mathrm{i}\right)\mathrm{becomes} \\ y=co{t}^{-1}\left(cot\frac{x}{2}\right) \\ \Rightarrow y=\frac{x}{2} \\ \therefore \frac{dy}{dx}=\frac{1}{2} \\ \mathrm{Hence},\frac{d\mathrm{y}}{d\mathrm{x}}\mathrm{is}\mathrm{independent}\mathrm{of}x. \end{gathered}$$