Prove that \( \tan^{-1}\left(\frac{2ab}{a^2-b^2}\right) + \tan^{-1}\left(\frac{2xy}{x^2-y^2}\right) = \tan^{-1}\left(\frac{2\alpha\beta}{\alpha^2-\beta^2}\right) \)
where \( \alpha = ax – by,\; \beta = ay + bx \)

Solution:

Use identity:

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

Rewrite:

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

\[ \tan^{-1}\left(\frac{2xy}{x^2-y^2}\right) = 2\tan^{-1}\left(\frac{y}{x}\right) \]

So LHS becomes:

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

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

Using identity:

\[ \tan^{-1}p + \tan^{-1}q = \tan^{-1}\left(\frac{p+q}{1 – pq}\right) \]

\[ = \tan^{-1}\left(\frac{\frac{b}{a} + \frac{y}{x}}{1 – \frac{by}{ax}}\right) \]

\[ = \tan^{-1}\left(\frac{bx + ay}{ax – by}\right) \]

Thus,

\[ \text{LHS} = 2\tan^{-1}\left(\frac{bx + ay}{ax – by}\right) \]

Now use identity again:

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

Let

\[ t = \frac{\beta}{\alpha} \]

Then,

\[ \text{LHS} = \tan^{-1}\left(\frac{2\alpha\beta}{\alpha^2 – \beta^2}\right) \]

which equals RHS.

Final Answer:

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