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Mathematics Question on Inverse Trigonometric Functions

Prove cot(1+sinx+1−sinx1+sinx−1−sinx=x2,x∈[0,π4](\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}=\frac{x}{2},x∈[0,\frac{π}{4}]

Answer

consider(1+sinx+1−sinx1+sinx−1−sinx(\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}

=((1+sinx+1−sinx)21+sinx)2−1−sinx)2=(\frac{(\sqrt{1+sinx}+\sqrt{1-sinx)^2}}{\sqrt{1+sinx)^2}-\sqrt{1-sinx)^2}}

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

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

=21+√1−sin2x)2sinx=1+cosxsinx2\frac{1+√1-sin^2x)}{2sin x}=\frac{1+cosx}{sin x}=2cos2x22sinx2cosx2\frac{2cos^2\frac{x}{2}}{2sin^\frac{x}{2}cos\frac{x}{2}}

cotx2cot\frac{x}{2}

=L.H.S=cot-1 (√1+sinx+√1−sinx√1+sinx−√1−sinx(\frac{√1+sinx+√1-sinx}{√1+sinx-√1-sinx} =cot-1 (cotx2)(cot\frac{x}{2})= R.H.S