Solveeit Logo
Login

Question

Mathematics Question on Inverse Trigonometric Functions

Prove tan−1x=12cos−1(1−x1+x),x∈[0,1]tan^{-1}\sqrt{x}=\frac{1}{2}cos^{-1}(\frac{1-x}{1+x}),x∈[0,1]

Answer

Let x=tan2θ.x=tan^2θ.
Then x=tanθ.=>θ=tan−1x.\sqrt{x}=tanθ. =>θ=tan^{-1}\sqrt{x}.
so 1−x1+x\frac {1-x}{1+x} = 1−tan2θ1+tan2θ\frac{1-tan^2θ}{1+tan^2θ} =cos2θ.
Now we have,
RHS=12cos⁡−11−x1+x=12cos⁡−1(cos⁡2θ)=12×2θ=θ=tan⁡−1x\frac12 \cos^{-1}\frac{1-x}{1+x} = \frac12\cos^{-1}(\cos2\theta)= \frac12\times2\theta=\theta=\tan^{-1}\sqrt{x}=LHS.