Trigonometrical expansion

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.

Absolutely brilliant sir -- however I did this in this way --

tan x = n tan y ;

So e^ix - e^-ixe^ix + e^-ix = ne^iy - ne^-iye^iy + e^-iy

Applying componendo and dividendo ,

we get 2e^ix2e^-ix = ( 1 + n ) e^iy + ( 1 - n ) e^-iy( 1 - n ) e^iy + ( 1 + n ) e^-iy

= ( 1+ n ) e^iy ( 1 + m e^-2iy )( 1 + n ) e^-iy ( 1 + m e^2iy ) ..............where m = 1 + n1 - n

Or , e^2ix = e^2iy ( 1 + m e^{-2iy )}1 + m e^2iy

Taking log on both sides ,

2 ix = 2 iy + log ( 1 + m e^-2iy ) - log ( 1 + m e^2iy )

Expanding the " log " - s and grouping the imaginary parts gives the answer .