frac{1 + sin A - cos A}{1 + sin A + cos A} = | vedclass

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Question

(C) Given expression: $\frac{1 + \sin A - \cos A}{1 + \sin A + \cos A}$Using the identities $1 - \cos A = 2 \sin^2 \frac{A}{2}$,$1 + \cos A = 2 \cos^2

Vedclass · Accepted Answer

$	an \frac{A}{2}$

Vedclass · Answer

(C) Given expression: $\frac{1 + \sin A - \cos A}{1 + \sin A + \cos A}$Using the identities $1 - \cos A = 2 \sin^2 \frac{A}{2}$,$1 + \cos A = 2 \cos^2 \frac{A}{2}$,and $\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$:$= \frac{2 \sin^2 \frac{A}{2} + 2 \sin \frac{A}{2} \cos \frac{A}{2}}{2 \cos^2 \frac{A}{2} + 2 \sin \frac{A}{2} \cos \frac{A}{2}}$$= \frac{2 \sin \frac{A}{2} (\sin \frac{A}{2} + \cos \frac{A}{2})}{2 \cos \frac{A}{2} (\cos \frac{A}{2} + \sin \frac{A}{2})}$$= \frac{\sin \frac{A}{2}}{\cos \frac{A}{2}} = 	an \frac{A}{2}$Trick: Put $A = 60^\circ$.$\frac{1 + \sin 60^\circ - \cos 60^\circ}{1 + \sin 60^\circ + \cos 60^\circ} = \frac{1 + \frac{\sqrt{3}}{2} - \frac{1}{2}}{1 + \frac{\sqrt{3}}{2} + \frac{1}{2}} = \frac{\frac{1}{2} + \frac{\sqrt{3}}{2}}{\frac{3}{2} + \frac{\sqrt{3}}{2}} = \frac{1 + \sqrt{3}}{3 + \sqrt{3}} = \frac{1 + \sqrt{3}}{\sqrt{3}(\sqrt{3} + 1)} = \frac{1}{\sqrt{3}}$Since $	an \frac{60^\circ}{2} = 	an 30^\circ = \frac{1}{\sqrt{3}}$,the correct option is $(C)$.

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

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