MCQ
જો $\sin \alpha=-\frac{3}{5},\pi < \alpha < \frac{3\pi}{2}$ તો $\cos \frac{\alpha}{2}=.........$
- A$\frac{-3}{\sqrt{10}}$
- ✓$\frac{-1}{\sqrt{10}}$
- C$\frac{1}{\sqrt{10}}$
- D$\frac{3}{\sqrt{10}}$
$\sin \alpha=-\frac{3}{5}$ અને $\pi < \alpha < \frac{3\pi}{2}$
$\Rightarrow \cos \alpha=-\sqrt{1-\sin^2\alpha}=-\sqrt{1-\frac{9}{25}}=-\sqrt{\frac{16}{25}}=-\frac{4}{5}$
હવે, $\cos \frac{\alpha}{2}=-\sqrt{\frac{1+\cos \alpha}{2}}\ \ \left(\because \pi < \alpha < \frac{3\pi}{2}\Rightarrow \frac{\pi}{2} < \frac{\alpha}{2} < \frac{3\pi}{4}\right)$
$=-\sqrt{\frac{1-\frac{4}{5}}{2}}=-\sqrt{\frac{5-4}{2×5}}=\frac{-1}{\sqrt{10}}$
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