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

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Question

(c) $\frac{{\sqrt 2 - \sin \alpha - \cos \alpha }}{{\sin \alpha - \cos \alpha }}$

$= \frac{{\sqrt 2 - \sqrt 2 \left\{ {\frac{1}{{\sqrt 2 }}\sin \al

Vedclass · Accepted Answer

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

Vedclass · Answer

(c) $\frac{{\sqrt 2 - \sin \alpha - \cos \alpha }}{{\sin \alpha - \cos \alpha }}$

$= \frac{{\sqrt 2 - \sqrt 2 \left\{ {\frac{1}{{\sqrt 2 }}\sin \alpha + \frac{1}{{\sqrt 2 }}\cos \alpha } ight\}}}{{\sqrt 2 \left\{ {\frac{1}{{\sqrt 2 }}\sin \alpha - \frac{1}{{\sqrt 2 }}\cos \alpha } ight\}}}$

$=\frac{{\sqrt 2 - \sqrt 2 \cos \left( {\alpha - \frac{\pi }{4}} ight)}}{{\sqrt 2 \sin \left( {\alpha - \frac{\pi }{4}} ight)}}$

$= \frac{{\sqrt 2 \left\{ {\,1 - \cos 	heta } ight\}}}{{\sqrt 2 \sin 	heta }},$  जहाँ  $	heta = \alpha - \frac{\pi }{4}$

$= \frac{{2{{\sin }^2}(	heta /2)}}{{2\sin (	heta /2)\cos (	heta /2)}} = 	an \frac{	heta }{2}$ 

$ = 	an \left( {\frac{\alpha }{2} - \frac{\pi }{8}} ight)$.

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

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