Evaluate: $\sqrt{3} \operatorname{cosec} 20^{\circ} - \sec 20^{\circ}$

$\cos ^2 5^{\circ}-\cos ^2 15^{\circ}-\sin ^2 15^{\circ}+\sin ^2 35^{\circ}+\cos 15^{\circ} \sin 15^{\circ}-\cos 5^{\circ} \sin 35^{\circ} = $

Assertion $(A)$: If $\sqrt{4 \sin^4 \theta + \sin^2 2\theta} + 4 \cos^2\left(\frac{\pi}{4} - \frac{\theta}{2}\right) = 2$,then $\theta$ lies in the $3^{\text{rd}}$ quadrant or $4^{\text{th}}$ quadrant.
Reason $(R)$: $\sqrt{\sin^2 \theta} = \sin \theta$

If $A = \sin \theta |\sin \theta|$,$B = \cos \theta |\cos \theta|$ and $\frac{99 \pi}{2} \leq \theta \leq \frac{100 \pi}{2}$,then

If $\frac{\cos (\theta_1+\theta_2)}{\cos (\theta_1-\theta_2)}+\frac{\cos (\theta_3-\theta_4)}{\cos (\theta_3+\theta_4)}=0$,then $\cot \theta_1 \cdot \cot \theta_2 \cdot \cot \theta_3 \cdot \cot \theta_4=$

The value of $(1-\cos \theta)(1+\cos \theta)(1+\cot^2 \theta)$,when $\theta = \frac{\pi}{15}$ is