For $A = 133^\circ ,\;2\cos \frac{A}{2}$ is equal to
For $A = 133^\circ ,\;2\cos \frac{A}{2}$ is equal to
- A
$ - \sqrt {1 + \sin A} - \sqrt {1 - \sin A} $
- B
$ - \sqrt {1 + \sin A} + \sqrt {1 - \sin A} $
- C
$\sqrt {1 + \sin A} - \sqrt {1 - \sin A} $
- D
$\sqrt {1 + \sin A} + \sqrt {1 - \sin A} $
Similar Questions
${\sin ^2}\frac{\pi }{8} + {\sin ^2}\frac{{3\pi }}{8} + {\sin ^2}\frac{{5\pi }}{8} + {\sin ^2}\frac{{7\pi }}{8} = $
${\sin ^2}\frac{\pi }{8} + {\sin ^2}\frac{{3\pi }}{8} + {\sin ^2}\frac{{5\pi }}{8} + {\sin ^2}\frac{{7\pi }}{8} = $
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]
The value of $\frac{{\tan {{70}^o} - \tan {{20}^o}}}{{\tan {{50}^o}}} = $
The value of $\frac{{\tan {{70}^o} - \tan {{20}^o}}}{{\tan {{50}^o}}} = $
If $\sin \theta+\cos \theta=\frac{1}{2}$, then $16(\sin (2 \theta)+\cos (4 \theta)+\sin (6 \theta))$ is equal to:
If $\sin \theta+\cos \theta=\frac{1}{2}$, then $16(\sin (2 \theta)+\cos (4 \theta)+\sin (6 \theta))$ is equal to:
- [JEE MAIN 2021]
If $\cos \,(\theta - \alpha ) = a,\,\,\sin \,(\theta - \beta ) = b,\,\,$then ${\cos ^2}(\alpha - \beta ) + 2ab\,\sin \,(\alpha - \beta )$ is equal to
If $\cos \,(\theta - \alpha ) = a,\,\,\sin \,(\theta - \beta ) = b,\,\,$then ${\cos ^2}(\alpha - \beta ) + 2ab\,\sin \,(\alpha - \beta )$ is equal to