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Trigonometry – II — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II2 Marks
MCQ
The value of the expression
$\frac{1+\sin 2 \alpha}{\cos (2 \alpha-2 \pi) \tan \left(\alpha-\frac{3 \pi}{4}\right)} -\frac{1}{4} \sin 2 \alpha\left(\cot \frac{\alpha}{2}+\cot \left(\frac{3 \pi}{2}+\frac{\alpha}{2}\right)\right)$ is
  • A
    $0$
  • B
    1
  • C
    $\sin ^2 \frac{\alpha}{2}$
  • ✓
    $\sin ^2 \alpha$
✓

Answer

Correct option: D.
$\sin ^2 \alpha$
(D)
$\frac{1+\sin 2 \alpha}{\cos (2 \alpha-2 \pi) \tan \left(\alpha-\frac{3 \pi}{4}\right)}$$-\frac{1}{4} \sin 2 \alpha\left(\cot \frac{\alpha}{2}+\cot \left(\frac{3 \pi}{2}+\frac{\alpha}{2}\right)\right)$
$=\frac{1+2 \sin \alpha \cos \alpha}{\cos 2 \alpha\left(\frac{\tan \alpha+1}{1-\tan \alpha}\right)}$$-\frac{1}{4}(2 \sin \alpha \cos \alpha)\left(\frac{\cos ^2 \frac{\alpha}{2}-\sin ^2 \frac{\alpha}{2}}{\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}\right)$
$=\frac{(\cos \alpha+\sin \alpha)^2}{\cos 2 \alpha\left(\frac{\cos \alpha+\sin \alpha}{\cos \alpha-\sin \alpha}\right)}-\frac{1}{4}(2 \sin \alpha \cos \alpha)\left(2 \frac{\cos \alpha}{\sin \alpha}\right)$
$=1-\cos ^2 \alpha=\sin ^2 \alpha$

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