If $\tan \beta = \cos \theta \tan \alpha ,$ then ${\tan ^2}\frac{\theta }{2} = $
If $\tan \beta = \cos \theta \tan \alpha ,$ then ${\tan ^2}\frac{\theta }{2} = $
- A
$\frac{{\sin (\alpha + \beta )}}{{\sin (\alpha - \beta )}}$
- B
$\frac{{\cos (\alpha - \beta )}}{{\cos (\alpha + \beta )}}$
- C
$\frac{{\sin (\alpha - \beta )}}{{\sin (\alpha + \beta )}}$
- D
$\frac{{\cos (\alpha + \beta )}}{{\cos (\alpha - \beta )}}$
Similar Questions
If $\sin 6\theta = 32{\cos ^5}\theta \sin \theta - 32{\cos ^3}\theta \sin \theta + 3x,$ then $x = $
If $\sin 6\theta = 32{\cos ^5}\theta \sin \theta - 32{\cos ^3}\theta \sin \theta + 3x,$ then $x = $
If $\sin A + \cos A = \sqrt 2 ,$ then ${\cos ^2}A = $
If $\sin A + \cos A = \sqrt 2 ,$ then ${\cos ^2}A = $
${\sin ^4}\frac{\pi }{8} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8} = $
${\sin ^4}\frac{\pi }{8} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8} = $
$\tan \alpha + 2\tan 2\alpha + 4\tan 4\alpha + 8\cot \,8\alpha = $
$\tan \alpha + 2\tan 2\alpha + 4\tan 4\alpha + 8\cot \,8\alpha = $
- [IIT 1988]
Given that $\cos \left( {\frac{{\alpha - \beta }}{2}} \right) = 2\cos \left( {\frac{{\alpha + B}}{2}} \right)$, then $\tan \frac{\alpha }{2}\tan \frac{\beta }{2} $ is equal to
Given that $\cos \left( {\frac{{\alpha - \beta }}{2}} \right) = 2\cos \left( {\frac{{\alpha + B}}{2}} \right)$, then $\tan \frac{\alpha }{2}\tan \frac{\beta }{2} $ is equal to