# Question #09530

Dec 31, 2016

To prove
$2 {\tan}^{-} 1 \left(\tan \left(\frac{A}{2}\right) \cdot \tan \left(\frac{\pi}{4} - \frac{B}{2}\right)\right) = {\tan}^{-} 1 \left(\frac{\sin A \cos B}{\cos A + \sin B}\right)$

Let $\frac{A}{2} = x \mathmr{and} \left(\frac{\pi}{4} - \frac{B}{2}\right) = y$
so
$2 x = A \mathmr{and} 2 y = \frac{\pi}{2} - B$

Now

$L H S = 2 {\tan}^{-} 1 \left(\tan \left(\frac{A}{2}\right) \cdot \tan \left(\frac{\pi}{4} - \frac{B}{2}\right)\right)$

$= 2 {\tan}^{-} 1 \left(\tan x \cdot \tan y\right)$

$= {\tan}^{-} 1 \left(\frac{2 \tan x \cdot \tan y}{1 - {\tan}^{2} x \cdot {\tan}^{2} y}\right)$

$= {\tan}^{-} 1 \left(\frac{2 \tan x \cdot \tan y \cdot {\cos}^{2} x \cdot {\cos}^{2} y}{{\cos}^{2} x {\cos}^{2} y - {\tan}^{2} x {\tan}^{2} y \cdot {\cos}^{2} x {\cos}^{2} y}\right)$

$= {\tan}^{-} 1 \left(\frac{2 \sin x \cos x \cdot 2 \sin y \cos y}{2 \left({\cos}^{2} x {\cos}^{2} y - {\sin}^{2} x {\sin}^{2} y\right)}\right)$

$= {\tan}^{-} 1 \left(\frac{\sin 2 x \sin 2 y}{2 \left(\cos x \cos y - \sin x \sin y\right) \left(\cos x \cos y + \sin x \sin y\right)}\right)$

$= {\tan}^{-} 1 \left(\frac{\sin 2 x \sin 2 y}{2 \cos \left(x + y\right) \cos \left(x - y\right)}\right)$

$= {\tan}^{-} 1 \left(\frac{\sin 2 x \sin 2 y}{\cos \left(2 x\right) + \cos \left(2 y\right)}\right)$

$= {\tan}^{-} 1 \left(\frac{\sin A \sin \left(\frac{\pi}{2} - B\right)}{\cos A + \cos \left(\frac{\pi}{2} - B\right)}\right)$

$= {\tan}^{-} 1 \left(\frac{\sin A \cos B}{\cos A + \sin B}\right) = R H S$

Proved