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Let $S=\left\{x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right): 9^{1-\tan ^2 x}+9^{\tan ^2 x}=10\right\}$ and $\beta=\sum_{x \in S} \tan ^2\left(\frac{x}{3}\right)$, then $\frac{1}{6}(\beta-14)^2$ is equal to
A
$32$
B
$8$
C
$64$
D
$16$
(JEE MAIN-2023)
Solution
Let $9^{\tan ^2 x}= P$
$\frac{9}{ P }+ P =10$
$P^2-10 P+9=0$
$(P-9)(P-1)=0$
$P=1,9$
$9^{\tan ^2 x}=1,9^{\tan ^2 x}=9$
$\tan ^2 x =0, \tan ^2 x =1$
$x =0, \pm \frac{\pi}{4} \quad \therefore x \in\left(-\frac{\pi}{2}, \frac{ p }{2}\right)$
$\beta=\tan ^2(0)+\tan ^2\left(+\frac{\pi}{12}\right)+\tan ^2\left(-\frac{\pi}{12}\right)$
$=0+2\left(\tan 15^{\circ}\right)^2$
$2(2-\sqrt{3})^2$
$2(7-4 \sqrt{3})$
Than $\frac{1}{6}(14-8 \sqrt{3}-14)^2=32$
Standard 11
Mathematics
