$$
2\tan ^{-1}\left(\sqrt{\frac{a}{b}}\tan \frac{x}{2}\right)=\sin ^{-1}\frac{2\sqrt{ab}\sin x}{\left(b+a\right)+\left(b-a\right)\cos x}
$$
I know within inverse of trigonometric function we have the value.
How Do I solve this.
my approach for this solution was:
1st:
$$
\sin ^{-1}\dfrac{2\sqrt{\dfrac{a}{b}}\tan \dfrac{x}{2}}{1+\left( \sqrt{\dfrac{a}{b}}\tan \dfrac{x}{2}\right) ^{2}}
$$
$$
\sin ^{-1}\dfrac{2\dfrac{\sqrt{a}}{\sqrt{b}}\tan \dfrac{x}{2}}{1+\dfrac{a}{b}\tan ^{2}\dfrac{n}{2}}
$$
$$
\sin \dfrac{2\sqrt{a}\tan \dfrac{x}{2}}{\dfrac{\sqrt{b}\left( b+atan ^{2}\dfrac{x}{2}\right) }{b}}
$$
$$
\sin ^{-1}\dfrac{2\sqrt{a}\tan \dfrac{x}{2}\sqrt{b}}{b+atan ^{2}\dfrac{x}{2}}
$$
$$
\sin ^{-1}\dfrac{2\sqrt{ab}\tan \dfrac{x}{2}}{b+atan ^{2}\dfrac{x}{2}}
$$
$$
\sin ^{-1}\dfrac{2\sqrt{ab}\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}}{b+a\dfrac{\sin ^{2}\dfrac{x}{2}}{\cos ^{2}\dfrac{x}{2}}}
$$
$$
=\sin ^{-1}\dfrac{2\sqrt{ab}\sin \dfrac{x}{2}\cos \dfrac{x}{2}}{\cos ^{2}\dfrac{x}{2}b+asin ^{2}\dfrac{x}{2}}
$$
After this I have tried to approach using
$$
\sin ^{2}\theta +\cos ^{2}\theta =1
$$
formula but could not reach the Right Hand Side.
Best Answer
This should be the next step of your attempt: $$\sin ^{-1}\dfrac{\sqrt{ab}\sin x}{b\left(\frac{1+\cos x}{2}\right)+a\left(\frac{1-\cos x}2\right)}$$ using half-angle formula.
The result follows immediately.