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

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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) $\frac{{\sqrt 2 - \sin \alpha - \cos \alpha }}{{\sin \alpha - \cos \alpha }}$

$= \frac{{\sqrt 2 - \sqrt 2 \left\{ {\frac{1}{{\sqrt 2 }}\sin \alpha + \frac{1}{{\sqrt 2 }}\cos \alpha } ight\}}}{{\sqrt 2 \left\{ {\frac{1}{{\sqrt 2 }}\sin \alpha - \frac{1}{{\sqrt 2 }}\cos \alpha } ight\}}}$

$=\frac{{\sqrt 2 - \sqrt 2 \cos \left( {\alpha - \frac{\pi }{4}} ight)}}{{\sqrt 2 \sin \left( {\alpha - \frac{\pi }{4}} ight)}}$

$= \frac{{\sqrt 2 \left\{ {\,1 - \cos 	heta } ight\}}}{{\sqrt 2 \sin 	heta }},$   where $	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 }} = $

Similar Questions

If $x = sec\, \phi - tan\, \phi$ & $y = cosec\, \phi + cot\, \phi$ then :

If $\tan \theta = \frac{{\sin \alpha - \cos \alpha }}{{\sin \alpha + \cos \alpha }},$ then $\sin \alpha + \cos \alpha $ and $\sin \alpha - \cos \alpha $ must be equal to

The value of $\frac{1}{4} \,\,tan \frac{\pi}{8} +\frac{1}{8} \,\,tan \frac{\pi}{16}+\frac{1}{16} \,\,tan \frac{\pi}{32}+.\,.\,.\,\infty $ terms is equal to-

The value of $cot\, 7\frac{{{1^0}}}{2}$ $+ tan\, 67 \frac{{{1^0}}}{2} - cot 67 \frac{{{1^0}}}{2} - tan7 \frac{{{1^0}}}{2}$ is :

If $a\tan \theta = b$, then $a\cos 2\theta + b\sin 2\theta = $