If $\alpha + \beta - \gamma = \pi ,$ then ${\sin ^2}\alpha + {\sin ^2}\beta - {\sin ^2}\gamma = $
If $\alpha + \beta - \gamma = \pi ,$ then ${\sin ^2}\alpha + {\sin ^2}\beta - {\sin ^2}\gamma = $
- [IIT 1980]
- A
$2\,\sin \alpha \,\sin \beta \,\cos \gamma $
- B
$ 2\,\cos \alpha \,\cos \beta \,\cos \gamma$
- C
$2\,\sin \alpha \,\sin \beta \sin \gamma $
- D
None of these
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${(\cos \alpha + \cos \beta )^2} + {(\sin \alpha + \sin \beta )^2} = $
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For any $\theta \, \in \,\left( {\frac{\pi }{4},\frac{\pi }{2}} \right)$, the expression $3\,{\left( {\sin \,\theta - \cos \,\theta } \right)^4} + 6{\left( {\sin \,\theta + \cos \,\theta } \right)^2} + 4\,{\sin ^6}\,\theta $ equals
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- [JEE MAIN 2019]
If $\tan \alpha = \frac{1}{7}$ and $\sin \beta = \frac{1}{{\sqrt {10} }}\left( {0 < \alpha ,\,\beta < \frac{\pi }{2}} \right)$, then $2\beta $ is equal to
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