If $\sin^{-1}\frac{2a}{1+a^2}-\cos^{-1}\frac{1-b^2}{1+b^2}=\tan^{-1}\frac{2x}{1-x^2}$ then what is value of x?

Question

If $\sin^{-1}\frac{2a}{1+a^2}-\cos^{-1}\frac{1-b^2}{1+b^2}=\tan^{-1}\frac{2x}{1-x^2}$ then what is value of x?
Solution
$\tan^{-1}x=\tan^{-1}a-\tan^{-1}b=\tan^{-1}\frac{a-b}{1+ab}$
x=$\frac{a-b}{1+ab}$

i have a doubt what is $\tan^{-1}a$ and $\tan^{-1}b$ and how come $\tan^{-1}x=\tan^{-1}a-\tan^{-1}b$?

Answer 1

Hint:

Assume a=tan(y) and similarly for x and b .
Solve the identity and replace y with $tan^{-1}$b .
You will get your expression.

Use this ,this and this from here for help with those identities.