$\frac{{\sin 3\theta - \cos 3\theta }}{{\sin \theta + \cos \theta }} + 1 = $
$\frac{{\sin 3\theta - \cos 3\theta }}{{\sin \theta + \cos \theta }} + 1 = $
- A
$2\sin 2\theta $
- B
$2 cos 2\theta$
- C
$\tan 2\theta $
- D
$\cot 2\theta $
Similar Questions
સાબિત કરો કે : $\cos ^{2} 2 x-\cos ^{2} 6 x=\sin 4 x \sin 8 x$
સાબિત કરો કે : $\cos ^{2} 2 x-\cos ^{2} 6 x=\sin 4 x \sin 8 x$
જો $\cos \,(\theta - \alpha ) = a,\,\,\sin \,(\theta - \beta ) = b,\,\,$ તો ${\cos ^2}(\alpha - \beta ) + 2ab\,\sin \,(\alpha - \beta ) = . . . .$
જો $\cos \,(\theta - \alpha ) = a,\,\,\sin \,(\theta - \beta ) = b,\,\,$ તો ${\cos ^2}(\alpha - \beta ) + 2ab\,\sin \,(\alpha - \beta ) = . . . .$
જો $\tan \alpha = \frac{1}{7},\;\tan \beta = \frac{1}{3},$ તો $\cos 2\alpha = $
જો $\tan \alpha = \frac{1}{7},\;\tan \beta = \frac{1}{3},$ તો $\cos 2\alpha = $
$cot 5^o$ -$tan5^o$ -$2$ $tan10^o$ -$4$ $tan 20^o$ -$8$ $cot40^o$ =
$cot 5^o$ -$tan5^o$ -$2$ $tan10^o$ -$4$ $tan 20^o$ -$8$ $cot40^o$ =
$\frac{{\cos A}}{{1 - \sin A}} = $
$\frac{{\cos A}}{{1 - \sin A}} = $