${\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{{\sqrt 2 - \sin \alpha - \cos \alpha }}{{\sin \alpha - \cos \alpha }} = $
$\frac{{\sqrt 2 - \sin \alpha - \cos \alpha }}{{\sin \alpha - \cos \alpha }} = $
જો $\frac{x}{{\cos \theta }} = \frac{y}{{\cos \left( {\theta - \frac{{2\pi }}{3}} \right)}} = \frac{z}{{\cos \left( {\theta + \frac{{2\pi }}{3}} \right)}},$ તો $x + y + z = $
જો $\frac{x}{{\cos \theta }} = \frac{y}{{\cos \left( {\theta - \frac{{2\pi }}{3}} \right)}} = \frac{z}{{\cos \left( {\theta + \frac{{2\pi }}{3}} \right)}},$ તો $x + y + z = $
જો $\frac{\sqrt{2} \sin \alpha}{\sqrt{1+\cos 2 \alpha}}=\frac{1}{7}$ અને $\sqrt{\frac{1-\cos 2 \beta}{2}}=\frac{1}{\sqrt{10}}$ $\alpha, \beta \in\left(0, \frac{\pi}{2}\right),$ તો $\tan (\alpha+2 \beta)$ મેળવો.
જો $\frac{\sqrt{2} \sin \alpha}{\sqrt{1+\cos 2 \alpha}}=\frac{1}{7}$ અને $\sqrt{\frac{1-\cos 2 \beta}{2}}=\frac{1}{\sqrt{10}}$ $\alpha, \beta \in\left(0, \frac{\pi}{2}\right),$ તો $\tan (\alpha+2 \beta)$ મેળવો.
- [JEE MAIN 2020]
જો $\frac{{5\pi }}{2} < x < 3\pi $,હોય તો $\frac{{\sqrt {1 - \sin x} + \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }}$ =
જો $\frac{{5\pi }}{2} < x < 3\pi $,હોય તો $\frac{{\sqrt {1 - \sin x} + \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }}$ =
$\left( {1 + \cos \frac{\pi }{9}} \right)\left( {1 + \cos \frac{{3\pi }}{9}} \right)\left( {1 + \cos \frac{{5\pi }}{9}} \right)\left( {1 + \cos \frac{{7\pi }}{9}} \right)$ની કિમત ............ થાય
$\left( {1 + \cos \frac{\pi }{9}} \right)\left( {1 + \cos \frac{{3\pi }}{9}} \right)\left( {1 + \cos \frac{{5\pi }}{9}} \right)\left( {1 + \cos \frac{{7\pi }}{9}} \right)$ની કિમત ............ થાય