We first prove the identity
\frac{\tan y(1-n)}{1+n \tan^2 y} = \frac{m\sin 2y}{1+m \cos 2y}; m = \frac{1-n}{1+n}
This is equivalent to proving that
\frac{\tan y}{1+n \tan^2 y} = \frac{\sin 2y}{(1+n) + (1-n) \cos 2y}
RHS = \frac{\sin 2y}{(1+n)+(1-n) \cos 2y} = \frac{\sin 2y}{2(\cos^2y + n \sin^2y)} = \frac{2 \sin y \cos y}{2(\cos^2y + n \sin^2y)}
Dividing Nr and Dr. by cos2y then shows that RHS = LHS
Notice that LHS is nothing but tan (y-x) as
\tan (y-x) = \frac{\tan y - \tan x}{1+ \tan x \tan y} = \frac{\tan y(1-n)}{1+n \tan^2 y}
Now, let z = m (\cos 2y + i \sin 2y) = m e^{i2y}
Then the expression
m \sin 2y - \frac{m^2 \sin 4y}{2} + \frac{m^3 \sin 6y}{3} -....
is equivalent to Im(\log (1+z))
Now 1+z = re^{i \theta} \Rightarrow \log (1+z) = \log r + i \theta
\Rightarrow Im(\log (1+z)) = \theta
and we have \tan \theta = \frac{m \sin 2y}{1+m \cos 2y}
For appropriate y, we have
\theta = \tan^{-1} \left(\frac{m \sin 2y}{1+m \cos 2y}\right)
Then \theta = \tan^{-1} \left(\frac{m \sin 2y}{1+m \cos 2y}\right) = y-x as proved above
Hence y-x = m \sin 2y - \frac{m^2 \sin 4y}{2} + \frac{m^3 \sin 3y}{3} -...
from which the given relation follows.
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