$\text{यदि } \sin(\alpha+\beta)=1, \sin(\alpha-\beta)=\frac{1}{2}, \alpha, \beta \in [0, \frac{\pi}{2}], \text{ तो } \tan(\alpha+2\beta) \cdot \tan(2\alpha+\beta) = ?$

$\frac{2\sin \theta \tan \theta (1 - \tan \theta ) + 2\sin \theta \sec^2 \theta}{(1 + \tan \theta)^2} = $

यदि $\tan^2 \alpha \tan^2 \beta + \tan^2 \beta \tan^2 \gamma + \tan^2 \gamma \tan^2 \alpha + 2\tan^2 \alpha \tan^2 \beta \tan^2 \gamma = 1$ है,तो $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$ का मान क्या है?

जब $\frac{\sin 9 \theta}{\cos 27 \theta}+\frac{\sin 3 \theta}{\cos 9 \theta}+\frac{\sin \theta}{\cos 3 \theta}=k(\tan 27 \theta-\tan \theta)$ परिभाषित है,तब $k=$

$\frac{1}{\sin 10^\circ} - \frac{\sqrt{3}}{\cos 10^\circ} =$

$0 < \theta < \frac{\pi}{2}$ के लिए,$\sum_{m=1}^6 \operatorname{cosec}\left(\theta+\frac{(m-1) \pi}{4}\right) \operatorname{cosec}\left(\theta+\frac{m \pi}{4}\right) = 4 \sqrt{2}$ के हल हैं: