Find the value of \(\tan\frac{1}{2}\bigg[\sin^{-1}\frac{2x}{1+x^2}+\cos^{-1}\frac{1-y^2}{1+y^2}\bigg],\mid x\mid<1,y>0\,and\:xy<1\)
Let x = tan θ.
Then, θ = \(\tan^{-1}x.\)
\(\sin^{-1}\frac{2x}{1}+x^2=\sin^{-1}(\frac{2\tan\theta}{1+\tan^2\theta})=\sin^{-1}(\sin^2\theta)=2\theta=2\tan^{-1}x\)
Let y = tan \(\Phi\). Then, \(\Phi\) = \(\tan^{-1}y.\)
\(\cos^{-1}\frac{1-y^2}{1+y^2}=\cos^{-1}(\frac{1-\tan^2\Phi}{1+\tan^2\Phi})=\cos^{-1}(cos^2\Phi)=2\Phi=2\tan^{-1}y\)
\(\therefore \tan\frac{1}{2}[\sin^{-1}\frac{2x}{1}+x^2+\cos^{-1}\frac{1-y^2}{1+y^2}]\)
= \(\tan\frac{1}{2}[2\tan^{-1}x+2\tan^{-1}y]\)
=\(\tan[tan^{-1}x+\tan{-1}y]\)
=\(\tan\bigg[\tan^{-1}(\frac {x+y}{1-xy})\bigg]\)
=\(\frac{x+y}{1-xy}\)
The elementary properties of inverse trigonometric functions will help to solve problems. Here are a few important properties related to inverse trigonometric functions:
Tan−1x + Tan−1y = π + tan−1 (x+y/ 1-xy), if xy > 1
Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1
Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1
Tan−1x + Tan−1y = -π + tan−1 (x+y/ 1-xy), if xy > 1
= x, if x∈[−π/2, π/2]
= π−x, if x∈[π/2, 3π/2]
=−2π+x, if x∈[3π/2, 5π/2] And so on.
= −x, ∈[−π,0]
= x, ∈[0,π]
= 2π−x, ∈[π,2π]
=−2π+x, ∈[2π,3π]
= x, (−π/2, π/2)
= x−π, (π/2, 3π/2)
= x−2π, (3π/2, 5π/2)