Question

Prove cot\((\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}=\frac{x}{2},x∈[0,\frac{π}{4}]\)

Answer 1

consider\((\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}\)

\(=(\frac{(\sqrt{1+sinx}+\sqrt{1-sinx)^2}}{\sqrt{1+sinx)^2}-\sqrt{1-sinx)^2}}\)

\(=(\frac{(\sqrt{1+sinx}+\sqrt{1-sinx)^2}}{(\sqrt{1+sinx)^2}-\sqrt{1-sinx)^2}}\)    ( by rationalizing)

\(=(\frac{1+sinx)+(1-sinx)+2√(1+sin x)(1-sin x)}{1+sinx-1+sin x}\)

=\(2\frac{1+√1-sin^2x)}{2sin x}=\frac{1+cosx}{sin x}\)=\(\frac{2cos^2\frac{x}{2}}{2sin^\frac{x}{2}cos\frac{x}{2}}\)

\(cot\frac{x}{2}\)

=L.H.S=cot^-1 \((\frac{√1+sinx+√1-sinx}{√1+sinx-√1-sinx}\) =cot^-1 \((cot\frac{x}{2})\)= R.H.S