frac{{sqrt 2 - sin alpha - cos alpha }}{{sin | vedclass

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Question

(C) Given expression: $\frac{{\sqrt 2 - \sin \alpha - \cos \alpha }}{{\sin \alpha - \cos \alpha }}$$= \frac{{\sqrt 2 - \sqrt 2 \left( {\frac{1}{{\sqrt

Vedclass · Accepted Answer

$	an \left( {\frac{\alpha }{2} - \frac{\pi }{8}} ight)$

Vedclass · Answer

(C) Given expression: $\frac{{\sqrt 2 - \sin \alpha - \cos \alpha }}{{\sin \alpha - \cos \alpha }}$$= \frac{{\sqrt 2 - \sqrt 2 \left( {\frac{1}{{\sqrt 2 }}\sin \alpha + \frac{1}{{\sqrt 2 }}\cos \alpha } ight)}}{{\sqrt 2 \left( {\frac{1}{{\sqrt 2 }}\sin \alpha - \frac{1}{{\sqrt 2 }}\cos \alpha } ight)}}$$= \frac{{\sqrt 2 - \sqrt 2 \cos \left( {\alpha - \frac{\pi }{4}} ight)}}{{\sqrt 2 \sin \left( {\alpha - \frac{\pi }{4}} ight)}}$Let $	heta = \alpha - \frac{\pi }{4}$. Then the expression becomes $\frac{{\sqrt 2 (1 - \cos 	heta )}}{{\sqrt 2 \sin 	heta }} = \frac{{1 - \cos 	heta }}{{\sin 	heta }}$Using half-angle identities $1 - \cos 	heta = 2 \sin^2(	heta/2)$ and $\sin 	heta = 2 \sin(	heta/2) \cos(	heta/2)$:$= \frac{{2{{\sin }^2}(	heta /2)}}{{2\sin (	heta /2)\cos (	heta /2)}} = 	an \frac{	heta }{2}$Substituting $	heta = \alpha - \frac{\pi }{4}$ back:$= 	an \left( {\frac{\alpha }{2} - \frac{\pi }{8}} ight)$.

$\frac{{\sqrt 2 - \sin \alpha - \cos \alpha }}{{\sin \alpha - \cos \alpha }} = $

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