Solve for $x$ from an equation containing inverse trigonometric functions

Question

How to solve the following for $x$? $$ \sin^{-1}\left(\frac{2a}{1+a^{2}}\right)+ \sin^{-1}\left(\frac{2b}{1+b^{2}}\right)= 2 \tan^{-1}(x ) $$

What conditions apply?

Answer 1

Let $\tan^{-1}a=A\implies -\dfrac\pi2\le A\le\dfrac\pi2\iff-\pi\le2A\le\pi$ and $a=\tan A$

$$\sin^{-1}\left[\dfrac{2\tan A}{1+\tan^2A}\right]=\sin^{-1}(\sin2A)$$

$$=\begin{cases} 2A &\mbox{if } -\dfrac\pi2\le2A\le\dfrac\pi2 \\ \pi-2A & \mbox{if } \dfrac\pi2<2A\le\pi \\ -\pi-2A & \mbox{if } -\pi\le2A<-\dfrac\pi2 \end{cases} $$

Answer 2

Hint: Let $\theta = \sin^{-1}\left(\dfrac{2a}{1+a^2}\right) \Rightarrow \sin \theta = \dfrac{2a}{1+a^2}$, and similarly $\sin \beta = \dfrac{2b}{1+b^2}$. Thus:

$\tan^{-1}x = \dfrac{\theta}{2}+\dfrac{\beta}{2} \Rightarrow x = \tan\left(\dfrac{\theta}{2}+\dfrac{\beta}{2}\right)$. You can compute $\tan\left(\dfrac{\theta}{2}\right)$, and $\tan\left(\dfrac{\beta}{2}\right)$ using the above equations, then find $x$.

Answer 3

$$ \sin^{-1}\left(\frac{2a}{1+a^{2}}\right)+ \sin^{-1}\left(\frac{2b}{1+b^{2}}\right)= 2 \tan^{-1}(x ) $$

Draw a right angled triangle with sides $$ (1+a^2), 2a, (1-a^2) $$ with all $ \tan^{-1}$

It now becomes

$$ \tan^{-1}\left(\frac{2a}{1-a^{2}}\right)+ \tan^{-1}\left(\frac{2b}{1-b^{2}}\right)= 2 \tan^{-1}(x ) $$

and proceed using double angle for tan as in the question you gave just alongside.