ધારોકે S=left{x inleft(-frac{pi}{2}, fra | vedclass

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Question

Let $9^{	an ^2 x}= P$

$\frac{9}{ P }+ P =10$

$P^2-10 P+9=0$

$(P-9)(P-1)=0$

$P=1,9$

$9^{	an ^2 x}=1,9^{	an ^2 x}=9$

$	an ^2 x =0,

Vedclass · Accepted Answer

$32$

Vedclass · Answer

Let $9^{	an ^2 x}= P$

$\frac{9}{ P }+ P =10$

$P^2-10 P+9=0$

$(P-9)(P-1)=0$

$P=1,9$

$9^{	an ^2 x}=1,9^{	an ^2 x}=9$

$	an ^2 x =0, 	an ^2 x =1$

$x =0, \pm \frac{\pi}{4} \quad 	herefore x \in\left(-\frac{\pi}{2}, \frac{ p }{2}ight)$

$\beta=	an ^2(0)+	an ^2\left(+\frac{\pi}{12}ight)+	an ^2\left(-\frac{\pi}{12}ight)$

$=0+2\left(	an 15^{\circ}ight)^2$

$2(2-\sqrt{3})^2$

$2(7-4 \sqrt{3})$

Than $\frac{1}{6}(14-8 \sqrt{3}-14)^2=32$

ધારોકે $S=\left\{x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right): 9^{1-\tan ^2 x}+9^{\tan ^2 x}=10\right\}$, અને $\beta=\sum_{x \in S} \tan ^2\left(\frac{x}{3}\right)$,તો $\frac{1}{6}(\beta-14)^2=.........$

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