$ \cos ^{3}\left(\frac{\pi}{8}\right) \cdot \cos \left(\frac{3 \pi}{8}\right)+\sin ^{3}\left(\frac{\pi}{8}\right) \cdot \sin \left(\frac{3 \pi}{8}\right)$ ની કિમંત મેળવો.
$\frac{1}{4}$
$\frac{1}{\sqrt{2}}$
$\frac{1}{2\sqrt{2}}$
$\frac{1}{2}$
$cosec \frac{\pi }{{18}} - \sqrt 3 \,sec\, \frac{\pi }{{18}}$ =
સાબિત કરો કે : $\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x$
જો $\alpha ,\,\beta ,\,\gamma \in \,\left( {0,\,\frac{\pi }{2}} \right)$, તો $\frac{{\sin \,(\alpha + \beta + \gamma )}}{{\sin \alpha + \sin \beta + \sin \gamma }} = . . ..$
$\frac{{\sin 3\theta - \cos 3\theta }}{{\sin \theta + \cos \theta }} + 1 = $
જો $\frac{{5\pi }}{2} < x < 3\pi $,હોય તો $\frac{{\sqrt {1 - \sin x} + \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }}$ =