Let S = {x in left(-frac{pi}{2}, frac{pi}{2}r | vedclass

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(A) Let $9^{	an^2 x} = P$.Given equation: $\frac{9}{P} + P = 10$.$P^2 - 10P + 9 = 0$.$(P - 9)(P - 1) = 0$.So,$P = 1$ or $P = 9$.Case $1$: $9^{	an^2

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(A) Let $9^{	an^2 x} = P$.Given equation: $\frac{9}{P} + P = 10$.$P^2 - 10P + 9 = 0$.$(P - 9)(P - 1) = 0$.So,$P = 1$ or $P = 9$.Case $1$: $9^{	an^2 x} = 1 \implies 	an^2 x = 0 \implies x = 0$.Case $2$: $9^{	an^2 x} = 9 \implies 	an^2 x = 1 \implies x = \pm \frac{\pi}{4}$.Thus,$S = \{0, \frac{\pi}{4}, -\frac{\pi}{4}\}$.$\beta = 	an^2(0) + 	an^2\left(\frac{\pi}{12}ight) + 	an^2\left(-\frac{\pi}{12}ight) = 0 + 2	an^2(15^{\circ})$.Since $	an(15^{\circ}) = 2 - \sqrt{3}$,we have $	an^2(15^{\circ}) = (2 - \sqrt{3})^2 = 7 - 4\sqrt{3}$.$\beta = 2(7 - 4\sqrt{3}) = 14 - 8\sqrt{3}$.Then $\frac{1}{6}(\beta - 14)^2 = \frac{1}{6}(14 - 8\sqrt{3} - 14)^2 = \frac{1}{6}(-8\sqrt{3})^2 = \frac{1}{6}(64 	imes 3) = \frac{192}{6} = 32$.

Let $S = \{x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) : 9^{1-\tan^2 x} + 9^{\tan^2 x} = 10\}$ and $\beta = \sum_{x \in S} \tan^2\left(\frac{x}{3}\right)$,then $\frac{1}{6}(\beta - 14)^2$ is equal to

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