MCQ
$\frac{{\sqrt 2 - \sin \alpha - \cos \alpha }}{{\sin \alpha - \cos \alpha }} = $
  • A
    $\sec \left( {\frac{\alpha }{2} - \frac{\pi }{8}} \right)$
  • B
    $\cos \left( {\frac{\pi }{8} - \frac{\alpha }{2}} \right)$
  • $\tan \left( {\frac{\alpha }{2} - \frac{\pi }{8}} \right)$
  • D
    $\cot \left( {\frac{\alpha }{2} - \frac{\pi }{2}} \right)$

Answer

Correct option: C.
$\tan \left( {\frac{\alpha }{2} - \frac{\pi }{8}} \right)$
c
(c) $\frac{{\sqrt 2 - \sin \alpha - \cos \alpha }}{{\sin \alpha - \cos \alpha }}$

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

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

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

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

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

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