જો $\tan \theta = t,$ તો $\tan 2\theta + \sec 2\theta = $
જો $\tan \theta = t,$ તો $\tan 2\theta + \sec 2\theta = $
- A
$\frac{{1 + t}}{{1 - t}}$
- B
$\frac{{1 - t}}{{1 + t}}$
- C
$\frac{{2t}}{{1 - t}}$
- D
$\frac{{2t}}{{1 + t}}$
Similar Questions
જો $\cos \left( {\alpha + \beta } \right) = \frac{4}{5}$ અને $\sin \left( {\alpha - \beta } \right) = \frac{5}{{13}}$,કે જ્યાં $0 \le \alpha ,\beta \le \frac{\pi }{4}$. તો $\tan 2\alpha $ મેળવો.
જો $\cos \left( {\alpha + \beta } \right) = \frac{4}{5}$ અને $\sin \left( {\alpha - \beta } \right) = \frac{5}{{13}}$,કે જ્યાં $0 \le \alpha ,\beta \le \frac{\pi }{4}$. તો $\tan 2\alpha $ મેળવો.
- [AIEEE 2010]
$\frac{{3 + \cot {{76}^o}\cot {{16}^o}}}{{\cot {{76}^o} + \cot {{16}^o}}}$ =
$\frac{{3 + \cot {{76}^o}\cot {{16}^o}}}{{\cot {{76}^o} + \cot {{16}^o}}}$ =
$\cos 2(\theta + \phi ) - 4\cos (\theta + \phi )\sin \theta \sin \phi + 2{\sin ^2}\phi = $
$\cos 2(\theta + \phi ) - 4\cos (\theta + \phi )\sin \theta \sin \phi + 2{\sin ^2}\phi = $
$\frac{{\sec 8A - 1}}{{\sec 4A - 1}} = $
$\frac{{\sec 8A - 1}}{{\sec 4A - 1}} = $
$\cos \frac{\pi }{7}\cos \frac{{2\pi }}{7}\cos \frac{{4\pi }}{7} = $
$\cos \frac{\pi }{7}\cos \frac{{2\pi }}{7}\cos \frac{{4\pi }}{7} = $