${\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} = $
- A
$1$
- B
$-1$
- C
$0$
- D
$2$
Similar Questions
$\frac{1}{{\sin 10^\circ }} - \frac{{\sqrt 3 }}{{\cos 10^\circ }} =$
$\frac{1}{{\sin 10^\circ }} - \frac{{\sqrt 3 }}{{\cos 10^\circ }} =$
- [IIT 1974]
यदि $\sin A + \cos A = \sqrt 2 ,$ तो ${\cos ^2}A = $
यदि $\sin A + \cos A = \sqrt 2 ,$ तो ${\cos ^2}A = $
$1 + \cos 2x + \cos 4x + \cos 6x = $
$1 + \cos 2x + \cos 4x + \cos 6x = $
यदि $\tan A = \frac{1}{2},\tan B = \frac{1}{3},$ तब $\cos 2A = $
यदि $\tan A = \frac{1}{2},\tan B = \frac{1}{3},$ तब $\cos 2A = $
यदि $\tan A = \frac{1}{2},$ तो $\tan 3A = $
यदि $\tan A = \frac{1}{2},$ तो $\tan 3A = $