The value of $\left(\sin 70^{\circ} ight)\left(\cot 10^{\circ} \cot 70^{\circ}-1 ight)$ is\

Question

The value of $\left(\sin 70^{\circ}ight)\left(\cot 10^{\circ} \cot 70^{\circ}-1ight)$ is\

Professor PC Thomas Classes · Answer

(b): We have, $$ \begin{aligned} & \sin 70^{\circ}\left(\frac{\cos 10^{\circ} \cos 70^{\circ}-\sin 10^{\circ} \sin 70^{\circ}}{\sin 10^{\circ} \sin 70^{\circ}}ight) \ &= \sin 70^{\circ}\left[\frac{\cos 80^{\circ}}{\sin 10^{\circ} \sin 70^{\circ}}ight] \ & {[\because \cos (A+B)=\cos A \cos B-\sin A \sin B] } \ &= \frac{\cos 80^{\circ}}{\cos 80^{\circ}}=1 \quad\left[\because \sin \left(90^{\circ}-	hetaight)=\cos 	hetaight] \end{aligned} $$