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(C) Given equation: $\cos 2x + a \sin x = 2a - 7$Using $\cos 2x = 1 - 2 \sin^2 x$,we get: $1 - 2 \sin^2 x + a \sin x = 2a - 7$$2 \sin^2 x - a \sin x +

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$48$

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(C) Given equation: $\cos 2x + a \sin x = 2a - 7$Using $\cos 2x = 1 - 2 \sin^2 x$,we get: $1 - 2 \sin^2 x + a \sin x = 2a - 7$$2 \sin^2 x - a \sin x + 2a - 8 = 0$$2(\sin^2 x - 4) - a(\sin x - 2) = 0$$2(\sin x - 2)(\sin x + 2) - a(\sin x - 2) = 0$$(\sin x - 2)(2 \sin x + 4 - a) = 0$Since $\sin x 
eq 2$,we must have $a = 2 \sin x + 4$.Since $-1 \leq \sin x \leq 1$,we have $a \in [2( -1) + 4, 2(1) + 4] = [2, 6]$.Thus,$p = 2$ and $q = 6$.Now,$r = 	an 9^{\circ} - 	an 27^{\circ} - 	an 27^{\circ} + 	an 81^{\circ} = (	an 9^{\circ} + \cot 9^{\circ}) - (	an 27^{\circ} + \cot 27^{\circ})$.Using $	an 	heta + \cot 	heta = \frac{2}{\sin 2	heta}$,we get $r = \frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}}$.Using $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\sin 54^{\circ} = \frac{\sqrt{5}+1}{4}$,we get $r = \frac{8}{\sqrt{5}-1} - \frac{8}{\sqrt{5}+1} = 8 \left( \frac{\sqrt{5}+1 - (\sqrt{5}-1)}{5-1} ight) = 8 \left( \frac{2}{4} ight) = 4$.Therefore,$pqr = 2 	imes 6 	imes 4 = 48$.

List-$A$	List-$B$
$(I)$ $\tan \left(\frac{\alpha + \beta}{2}\right) =$	$(a)$ $\frac{b}{a}$
$(II)$ $\cos (\alpha + \beta) =$	$(b)$ $\frac{2ab}{a^2 + b^2}$
$(III)$ $\sin (\alpha + \beta) =$	$(c)$ $\frac{2ab}{a^2 - b^2}$
$(IV)$ $\tan (\alpha + \beta) =$	$(d)$ $\frac{a^2 - b^2}{a^2 + b^2}$

Let the set of all $a \in \mathbb{R}$ such that the equation $\cos 2x + a \sin x = 2a - 7$ has a solution be $[p, q]$ and $r = \tan 9^{\circ} - \tan 27^{\circ} - \frac{1}{\cot 63^{\circ}} + \tan 81^{\circ}$,then $pqr$ is equal to ....................

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