यदि $\sin x + \cos x = \frac{1}{5},$ तब $\tan 2x$ का मान होगा
यदि $\sin x + \cos x = \frac{1}{5},$ तब $\tan 2x$ का मान होगा
- A
$\frac{{25}}{{17}}$
- B
$\frac{{7}}{{25}}$
- C
$\frac{{25}}{7}$
- D
$\frac{{24}}{7}$
Similar Questions
${\cos ^2}A{(3 - 4{\cos ^2}A)^2} + {\sin ^2}A{(3 - 4{\sin ^2}A)^2} = $
${\cos ^2}A{(3 - 4{\cos ^2}A)^2} + {\sin ^2}A{(3 - 4{\sin ^2}A)^2} = $
माना $\cos (\alpha+\beta)=\frac{4}{5}$ और $\sin (\alpha-\beta)=\frac{5}{13},$ जहाँ $0 \leq \alpha, \beta \leq \frac{\pi}{4}$ तो $\tan 2 \alpha$ बराबर है
माना $\cos (\alpha+\beta)=\frac{4}{5}$ और $\sin (\alpha-\beta)=\frac{5}{13},$ जहाँ $0 \leq \alpha, \beta \leq \frac{\pi}{4}$ तो $\tan 2 \alpha$ बराबर है
- [IIT 1979]
यदि $\tan \beta = \cos \theta \tan \alpha ,$ तब ${\tan ^2}\frac{\theta }{2} = $
यदि $\tan \beta = \cos \theta \tan \alpha ,$ तब ${\tan ^2}\frac{\theta }{2} = $
$\frac{1}{{\sin 10^\circ }} - \frac{{\sqrt 3 }}{{\cos 10^\circ }} =$
$\frac{1}{{\sin 10^\circ }} - \frac{{\sqrt 3 }}{{\cos 10^\circ }} =$
- [IIT 1974]
$\frac{{\sin 3\theta - \cos 3\theta }}{{\sin \theta + \cos \theta }} + 1 = $
$\frac{{\sin 3\theta - \cos 3\theta }}{{\sin \theta + \cos \theta }} + 1 = $