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  4. ( √2 - sin α - cos α / sinα - cosα ) is equal to

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Q. $ \frac{ \sqrt{2} - \sin \alpha - \cos \alpha }{\sin\alpha - \cos\alpha } $ is equal to

Trigonometric Functions

Solution:

$ \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} \left(\theta/2\right)}{2 \sin \left(\theta /2\right) \cos \left(\theta/2\right)} = \tan \left(\theta/2\right) $
$ = \tan\left(\frac{\alpha}{2} - \frac{\pi}{8}\right) $