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Mathematics Question on Inverse Trigonometric Functions

Find the value of tan⁡12[sin⁡−12x1+x2+cos⁡−11−y21+y2],∣x∣<1,y>0 and xy<1\tan\frac{1}{2}\bigg[\sin^{-1}\frac{2x}{1+x^2}+\cos^{-1}\frac{1-y^2}{1+y^2}\bigg],\mid x\mid<1,y>0\,and\:xy<1

Answer

Let x = tan θ.
Then, θ = tan⁡−1x.\tan^{-1}x.

sin⁡−12x1+x2=sin⁡−1(2tan⁡θ1+tan⁡2θ)=sin⁡−1(sin⁡2θ)=2θ=2tan⁡−1x\sin^{-1}\frac{2x}{1}+x^2=\sin^{-1}(\frac{2\tan\theta}{1+\tan^2\theta})=\sin^{-1}(\sin^2\theta)=2\theta=2\tan^{-1}x

Let y = tan Φ\Phi. Then, Φ\Phi = tan⁡−1y.\tan^{-1}y.

cos⁡−11−y21+y2=cos⁡−1(1−tan⁡2Φ1+tan⁡2Φ)=cos⁡−1(cos2Φ)=2Φ=2tan⁡−1y\cos^{-1}\frac{1-y^2}{1+y^2}=\cos^{-1}(\frac{1-\tan^2\Phi}{1+\tan^2\Phi})=\cos^{-1}(cos^2\Phi)=2\Phi=2\tan^{-1}y

∴tan⁡12[sin⁡−12x1+x2+cos⁡−11−y21+y2]\therefore \tan\frac{1}{2}[\sin^{-1}\frac{2x}{1}+x^2+\cos^{-1}\frac{1-y^2}{1+y^2}]

= tan⁡12[2tan⁡−1x+2tan⁡−1y]\tan\frac{1}{2}[2\tan^{-1}x+2\tan^{-1}y]

=tan⁡[tan−1x+tan⁡−1y]\tan[tan^{-1}x+\tan{-1}y]

=tan⁡[tan⁡−1(x+y1−xy)]\tan\bigg[\tan^{-1}(\frac {x+y}{1-xy})\bigg]

=x+y1−xy\frac{x+y}{1-xy}