If $\cos x = \frac{2 \cos y - 1}{2 - \cos y}$ where $x, y \in (0, \pi)$,then $\tan(x/2) \cot(y/2) =$

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 $(R)$: $\sqrt{\sin^2 \theta} = \sin \theta$

For a positive integer $n$,let ${f_n}(\theta ) = \left( {\tan \frac{\theta }{2}} \right)(1 + \sec \theta )(1 + \sec 2\theta )(1 + \sec 4\theta ) \dots (1 + \sec {2^n}\theta ).$ Then

If $\alpha \in (0, \pi/2)$,then $\sqrt{x^2 + x} + \frac{\tan^2 \alpha}{\sqrt{x^2 + x}} = \dots$

If $\sin \theta = \frac{2t}{1+t^2}$ and $\theta$ lies in the second quadrant, then $\cos \theta$ is equal to

If $\sin \alpha + \cos \alpha = m$,then $\sin^6 \alpha + \cos^6 \alpha = $