જો $\tan \beta = \cos \theta \tan \alpha ,$ તો ${\tan ^2}\frac{\theta }{2} = $
જો $\tan \beta = \cos \theta \tan \alpha ,$ તો ${\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
$ \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.
$ \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.
- [JEE MAIN 2020]
$\frac{{\tan {{70}^o} - \tan {{20}^o}}}{{\tan {{50}^o}}} = $
$\frac{{\tan {{70}^o} - \tan {{20}^o}}}{{\tan {{50}^o}}} = $
${\sin ^2}\frac{\pi }{8} + {\sin ^2}\frac{{3\pi }}{8} + {\sin ^2}\frac{{5\pi }}{8} + {\sin ^2}\frac{{7\pi }}{8} = $
${\sin ^2}\frac{\pi }{8} + {\sin ^2}\frac{{3\pi }}{8} + {\sin ^2}\frac{{5\pi }}{8} + {\sin ^2}\frac{{7\pi }}{8} = $
$tan^{-1} (\frac{sin2 -1}{cos2})$ =
$tan^{-1} (\frac{sin2 -1}{cos2})$ =
$\tan \frac{A}{2} = . . .$
$\tan \frac{A}{2} = . . .$