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The value of $\tan \frac{2\pi}{5} - \tan \frac{\pi}{15} - \sqrt{3} \tan \frac{2\pi}{5} \tan \frac{\pi}{15}$ is equal to

$\cos ^2\left(\frac{\pi}{6}+\theta\right)-\sin ^2\left(\frac{\pi}{6}-\theta\right)$ is equal to

If $\cot \alpha = 1$ and $\sec \beta = -\frac{5}{3}$,where $\pi < \alpha < \frac{3\pi}{2}$ and $\frac{\pi}{2} < \beta < \pi$,then the value of $\tan(\alpha + \beta)$ and the quadrant in which $\alpha + \beta$ lies,respectively,are

If $\frac{\pi}{2} < \alpha < \pi$ and $\pi < \beta < \frac{3\pi}{2}$,with $\sin \alpha = \frac{15}{17}$ and $\tan \beta = \frac{12}{5}$,then the value of $\sin(\beta - \alpha)$ is (in $/221$)

If $A=35^{\circ}, B=15^{\circ}$ and $C=40^{\circ}$,then $\tan A \cdot \tan B+\tan B \cdot \tan C+\tan C \cdot \tan A$ is equal to

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