If \( \sin^{-1}\left(\frac{2a}{1+a^2}\right) + \sin^{-1}\left(\frac{2b}{1+b^2}\right) = 2\tan^{-1}(x) \), prove that \( x = \frac{a+b}{1-ab} \)

Solution:

Use the identity:

\[ \sin^{-1}\left(\frac{2t}{1+t^2}\right) = 2\tan^{-1}(t) \]

Thus,

\[ \sin^{-1}\left(\frac{2a}{1+a^2}\right) = 2\tan^{-1}(a) \]

\[ \sin^{-1}\left(\frac{2b}{1+b^2}\right) = 2\tan^{-1}(b) \]

So the given equation becomes:

\[ 2\tan^{-1}(a) + 2\tan^{-1}(b) = 2\tan^{-1}(x) \]

Divide both sides by 2:

\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}(x) \]

Using identity:

\[ \tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \]

Hence,

\[ x = \frac{a+b}{1-ab} \]

Final Answer:

\[ x = \frac{a+b}{1-ab} \]