Question

What is the value of $\frac{\sqrt{2}-\sin \alpha -\cos \alpha }{\sin \alpha -\cos \alpha }?$

• A. $sec\left(\frac{\alpha }{2}-\frac{\mathrm{\pi }}{8}\right)$
• B. $\cos \left(\frac{\mathrm{\pi }}{8}-\frac{\alpha }{2}\right)$
• C. $\tan \left(\frac{\alpha }{2}-\frac{\mathrm{\pi }}{8}\right)$
• D. $cot\left(\frac{\alpha }{2}-\frac{\mathrm{\pi }}{2}\right)$

Answer 1

The correct option is C

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

Explanation for correct option:

Simplification of the given expression:

The given expression is: $\frac{\sqrt{2}-\sin \alpha -\cos \alpha }{\sin \alpha -\cos \alpha }$

Simplifying the above expression:

$$\begin{gathered} =\frac{\sqrt{2}-\sin \alpha -\cos \alpha }{\sin \alpha -\cos \alpha } \\ =\frac{\sqrt{2}-\sqrt{2}\left\{{\displaystyle \frac{1}{\sqrt{2}}}\sin \alpha +{\displaystyle \frac{1}{\sqrt{2}}}\cos \alpha \right\}}{\sqrt{2}\left\{{\displaystyle \frac{1}{\sqrt{2}}}\sin \alpha -{\displaystyle \frac{1}{\sqrt{2}}}\cos \alpha \right\}}\left[Multiplyanddivideby\sqrt{2}\right] \\ =\frac{\sqrt{2}-\sqrt{2}\left\{\sin {\displaystyle \frac{\mathrm{\pi }}{4}}\sin \alpha +\cos {\displaystyle \frac{\mathrm{\pi }}{4}}\cos \alpha \right\}}{\sqrt{2}\left\{\cos {\displaystyle \frac{\mathrm{\pi }}{4}}\sin \alpha -\sin {\displaystyle \frac{\mathrm{\pi }}{4}}\cos \alpha \right\}} \\ =\frac{\sqrt{2}-\sqrt{2}\cos \left(\alpha -{\displaystyle \frac{\mathrm{\pi }}{4}}\right)}{\sqrt{2}\sin \left(\alpha -{\displaystyle \frac{\mathrm{\pi }}{4}}\right)} \\ Let,\alpha -\frac{\mathrm{\pi }}{4}=\theta ----\left(\mathfrak{i}\right) \\ =\frac{\sqrt{2}-\sqrt{2}\cos \theta }{\sqrt{2}\sin \theta } \\ =\frac{\sqrt{2}\left(1-\cos \theta \right)}{\sqrt{2}\sin \theta } \\ =\frac{1-\cos \theta }{\sin \theta } \\ =\frac{2{\sin }^2\left({\displaystyle \frac{\theta }{2}}\right)}{2\sin \left({\displaystyle \frac{\theta }{2}}\right)\cos \left({\displaystyle \frac{\theta }{2}}\right)}\left[Trigonometricidentities\right] \\ =\frac{\sin \left({\displaystyle \frac{\theta }{2}}\right)}{\cos \left({\displaystyle \frac{\theta }{2}}\right)} \\ =\tan \left(\frac{\theta }{2}\right)\left[\because \frac{sin\theta }{cos\theta }=tan\theta \right] \\ =\tan \left(\frac{\alpha }{2}-\frac{\mathrm{\pi }}{8}\right)\mathbf{}\left[\theta =\alpha -\frac{\pi }{4}\right] \end{gathered}$$

Hence, the correct answer is optin (C).