Question

CosA/1-SinA=Tan(π/4+A/2)

Answer 1

use of two concepts
sin2A = 2sinA cosA

Answer 2

$\red{\frac{Cos A}{1 - SinA } } \\= \frac{ Cos^{2} \frac{A}{2} - Sin^{2} \frac{A}{2} } { 1 - 2 Sin \frac{A}{2} Cos \frac{A}{2}} \\= \frac{ (Cos \frac{A}{2} - Sin \frac{A}{2} )( Cos \frac{A}{2} +Sin \frac{A}{2})} { Cos^{2} \frac{A}{2} + Sin^{2} \frac{A}{2} - 2 Sin \frac{A}{2} Cos \frac{A}{2}}$

$= \frac{ (Cos \frac{A}{2} - Sin \frac{A}{2} )( Cos \frac{A}{2} +Sin \frac{A}{2})} { (Cos \frac{A}{2} - Sin\frac{A}{2} )^{2}}$

$= \frac{ ( Cos \frac{A}{2} +Sin \frac{A}{2})} { (Cos\frac{A}{2} - Sin\frac{A}{2} )}$

$Dividing \: numerator \:and \: denominator \\by \: Cos \frac{A}{2} , we \:get$

$= \frac{ 1 + tan \frac{A}{2}}{ 1 - tan \frac{A}{2}} \\= \frac{ tan \frac{\pi}{4} + tan \frac{A}{2}}{ 1 - tan \frac{\pi}{4} tan \frac{A}{2}} \\\green {= tan \Big( \frac{\pi}{4} + \frac{A}{2} \Big)} \\= RHS$

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