$ \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.
$\frac{1}{4}$
$\frac{1}{\sqrt{2}}$
$\frac{1}{2\sqrt{2}}$
$\frac{1}{2}$
$cot\, 7\frac{{{1^0}}}{2}$ $+ tan\, 67 \frac{{{1^0}}}{2} - cot 67 \frac{{{1^0}}}{2} - tan7 \frac{{{1^0}}}{2}$ =
$\frac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}} = $
$\cos \alpha .\sin (\beta - \gamma ) + \cos \beta .\sin (\gamma - \alpha ) + \cos \gamma .\sin (\alpha - \beta ) = $
$2\,{\sin ^2}\beta + 4\,\,\cos \,(\alpha + \beta )\,\,\sin \,\alpha \,\sin \,\beta + \cos \,2\,(\alpha + \beta ) = $
કોઈ પણ $\theta \, \in \,\left( {\frac{\pi }{4},\frac{\pi }{2}} \right)$ માટે, $3\,{\left( {\sin \,\theta - \cos \,\theta } \right)^4} + 6{\left( {\sin \,\theta + \cos \,\theta } \right)^2} + 4\,{\sin ^6}\,\theta $ =