In any triangle $ABC ,$ ${\sin ^2}\frac{A}{2} + {\sin ^2}\frac{B}{2} + {\sin ^2}\frac{C}{2}$ is equal to
In any triangle $ABC ,$ ${\sin ^2}\frac{A}{2} + {\sin ^2}\frac{B}{2} + {\sin ^2}\frac{C}{2}$ is equal to
- A
$1 - 2\,\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$
- B
$1 - 2\,\sin \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$
- C
$1 - 2\,\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$
- D
$1 - 2\,\cos \frac{A}{2}\cos \frac{B}{2}\sin \frac{C}{2}$
Similar Questions
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$\tan 75^\circ - \cot 75^\circ = $
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The expression $\frac{{{{\tan }^2}20^\circ - {{\sin }^2}20^\circ }}{{{{\tan }^2}20^\circ \,\cdot\,{{\sin }^2}20^\circ }}$ simplifies to
The expression $\frac{{{{\tan }^2}20^\circ - {{\sin }^2}20^\circ }}{{{{\tan }^2}20^\circ \,\cdot\,{{\sin }^2}20^\circ }}$ simplifies to
If $\cos \left( {\alpha + \beta } \right) = \frac{4}{5}$ and $\sin \left( {\alpha - \beta } \right) = \frac{5}{{13}}$,where $0 \le \alpha ,\beta \le \frac{\pi }{4}$ . Then $\tan 2\alpha =$
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- [IIT 1979]
$\frac{{\sqrt 2 - \sin \alpha - \cos \alpha }}{{\sin \alpha - \cos \alpha }} = $
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