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Trigonometry – II — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II2 Marks
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)
$\frac{\sqrt{2}-\sin \alpha-\cos \alpha}{\sin \alpha-\cos \alpha}$
$=\frac{\sqrt{2}-\sqrt{2}\left\{\frac{1}{\sqrt{2}} \sin \alpha+\frac{1}{\sqrt{2}} \cos \alpha\right\}}{\sqrt{2}\left\{\frac{1}{\sqrt{2}} \sin \alpha-\frac{1}{\sqrt{2}} \cos \alpha\right\}}$
$=\frac{\sqrt{2}-\sqrt{2} \cos \left(\alpha-\frac{\pi}{4}\right)}{\sqrt{2} \sin \left(\alpha-\frac{\pi}{4}\right)}$
$=\frac{\sqrt{2}(1-\cos \theta)}{\sqrt{2} \sin \theta}$, where $\theta=\alpha-\frac{\pi}{4}$
$=\frac{2 \sin ^2\left(\frac{\theta}{2}\right)}{2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}$
$=\tan \frac{\theta}{2}=\tan \left(\frac{\alpha}{2}-\frac{\pi}{8}\right) \quad \ldots\left[\because \theta=\alpha-\frac{\pi}{4}\right]$

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