Prove the following trigonometric identity:

$tan \left ( \dfrac{\pi}{4} - \dfrac{A}{2} \right ) = \dfrac{ cos \frac{A}{2} - sin\frac{A}{2} } {cos \frac{A}{2} + sin \frac{A}{2}} = \dfrac{cosA}{1+sinA}$

Solution:

Left Side

$= tan ( \frac{\pi}{4} - \frac{A}{2} )$

$= tan (\frac{180°}{4} - \frac{A}{2} )$

$= tan (45° - \frac{A}{2})$

$= \dfrac{tan45° - tan \frac{A}{2}}{tan45° + tan \frac{A}{2}}$

$= \dfrac{1 - tan\frac{A}{2} } {1+ tan \frac{A}{2}}$

$= \dfrac{ 1 - \dfrac{sin \frac{A}{2}}{cos \frac{A}{2}}}{1 + \dfrac{sin\frac{A}{2}}{cos \frac{A}{2}}}$

$= \dfrac{ \dfrac{cos \frac{A}{2} - sin\frac}A}{2} }{cos \frac{A}{2}} }{\dfrac{cos \frac{A}{2} + sin\frac{A}{2}}{cos \frac{A}{2}}}$

$= \dfrac{ cos\frac{A}{2} - sin\frac{A}{2}}{cos \frac{A}{2} + sin\frac{A}{2}}$

Middle Side

$= \dfrac{cos \frac{A}{2} - sin\frac{A}{2}}{cos \frac{A}{2} + sin\frac{A}{2}} × \dfrac{cos \frac{A}{2} + sin\frac{A}{2}}{cos \frac{A}{2} + sin\frac{A}{2}}$

$= \dfrac{(cos^2 \frac{A}{2} - sin^2 \frac{A}{2} )}{(cos^2 \frac{A}{2} + sin^2 \frac{A}{2}) +( 2sin \frac{A}{2} cos \frac{A}{2})}$

Contracting to formulae, we get,

$= \dfrac{cosA}{1 + sinA}$

Right Side