Question

Differentiate w.r.t. \(x\) the function: \(cot^{-1}\bigg[\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}\bigg],0<x<\frac{x}{2}\)

Answer 1

The correct answer is \(\frac{dy}{dx}=\frac{1}{2}\)
Let \(y=cot^{-1}\bigg[\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}\bigg]......(1)\)
Then,\(\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}\)
\(=\frac{(\sqrt{1+sinx}+\sqrt{1-sinx)^2}}{(\sqrt{1+sinx}-\sqrt{1-sinx})(\sqrt{1+sinx}+\sqrt{1-sinx})}\)
\(=\frac{(1+sinx)+(1-sinx)+2\sqrt{(1-sinx)(1+sinx)}}{(1+sinx)-(1-sinx)}\)
\(=\frac{2+2\sqrt{1-sin^2x}}{2sinx}\)
\(=\frac{1+cosx}{sinx}\)
\(=\frac{2cos^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}\)
\(=cot\frac{x}{2}\)
\(y=cot^{-1}(cot\frac{x}{2})\)
\(⇒y=\frac{x}{2}\)
\(∴\frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}(x)\)
\(⇒\frac{dy}{dx}=\frac{1}{2}\)