STD 11 - 3. trignometrical ratios,functions and identities — JEE कक्षा 11 साइन्स — Question

$= \frac{{\sqrt 2 - \sqrt 2 \left\{ {\frac{1}{{\sqrt 2 }}\sin \alpha + \frac{1}{{\sqrt 2 }}\cos \alpha } \right\}}}{{\sqrt 2 \left\{ {\frac{1}{{\sqrt 2 }}\sin \alpha - \frac{1}{{\sqrt 2 }}\cos \alpha } \right\}}}$

$=\frac{{\sqrt 2 - \sqrt 2 \cos \left( {\alpha - \frac{\pi }{4}} \right)}}{{\sqrt 2 \sin \left( {\alpha - \frac{\pi }{4}} \right)}}$

$= \frac{{\sqrt 2 \left\{ {\,1 - \cos \theta } \right\}}}{{\sqrt 2 \sin \theta }},$  जहाँ  $\theta = \alpha - \frac{\pi }{4}$

$= \frac{{2{{\sin }^2}(\theta /2)}}{{2\sin (\theta /2)\cos (\theta /2)}} = \tan \frac{\theta }{2}$ 

$ = \tan \left( {\frac{\alpha }{2} - \frac{\pi }{8}} \right)$.