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Mathematics Question on Inverse Trigonometric Functions

Prove tan-1(√1+x−√1−x√1+x+√1−x=π4−12cos−1x,−1√2≤x≤1.(\frac{√1+x-√1-x}{√1+x+√1-x}=\frac{π}{4}-\frac{1}{2}cos^-1x,-\frac{1}{√2}≤x≤1.[Hint: putx = cos 2θ]

Answer

Put x=cos2θ so that θ=12cos−1\frac{1}{2}cos-1x.Then we have:
LHS=tan−1(√1+x−√1−x√1+x+√1−x)LHS=tan^-1(\frac{√1+x-√1-x}{√1+x+√1-x)}
=tan−1(√1+cos2θ−√1−cos2θ√1+cos2θ+√1−cos2θ)=tan^-1(\frac{√1+cos2θ-√1-cos2θ}{√1+cos2θ+√1-cos2θ})
=tan−1(√2cos2θ−√2sin2θ√2cos2θ+√2sin2θ)=tan^-1(\frac{√2cos^2θ-√2sin^2θ}{√2cos^2θ+√2sin^2θ)}
=tan−1(√2cosθ−√2sinθ√2cosθ+√2sinθ)=tan^-1(\frac{√2cosθ-√2sinθ}{√2cosθ+√2sinθ})
=tan−1(cosθ−sinθcosθ+sinθ)=tan−1(1−tanθ1+tanθ)=tan^-1(\frac{cosθ-sinθ}{cosθ+sinθ})=tan-1(\frac{1-tanθ}{1+tanθ})
=tan−1(1)−tan−1(tanθ)=π4−θ=π4−12cos−1x=RHS=tan^-1(1)-tan^-1(tanθ)= \frac{π}{4}-θ=\frac{π}{4}-\frac{1}{2}cos^-1x=RHS