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

Solution:

Use standard inverse trigonometric identities:

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

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

So the given equation becomes:

\[ 2\tan^{-1}(a) – 2\tan^{-1}(b) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \]

\[ \Rightarrow 2\left[\tan^{-1}(a) – \tan^{-1}(b)\right] = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \]

Using identity:

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

\[ \Rightarrow 2\tan^{-1}\left(\frac{a-b}{1+ab}\right) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \]

Now use identity:

\[ \tan(2\theta) = \frac{2\tan\theta}{1 – \tan^2\theta} \]

This implies:

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

Comparing both sides, we get:

\[ t = x \]

Where

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

Hence,

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

Final Answer:

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