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  • Let <img alt="S=\left\{x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right): 9^{1-\tan ^2 x}+9^{\tan ^2 x}=10\right\}" src="https://entrancecorner.oncodecogs.com/gif.latex?S%3D%5Cleft%5C%7Bx%20%5Cin%5Cl

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  \mathrm{b}=\sum_{\mathrm{x} \in \mathrm{S}} \tan ^2\left(\frac{\mathrm{x}}{3}\right)$, then $\frac{1}{6}(\beta-14)^2is equal to

Option: 1

16


Option: 2

32


Option: 3

8


Option: 4

64


Answers (1)

best_answer

Let 9^{\tan ^2 x}=P

\begin{aligned} & \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 \end{aligned}

\begin{aligned} & 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(\tan 15)^2 \\ & 2(2-\sqrt{3})^2 \\ & 2(7-4 \sqrt{3}) \end{aligned}

Then  \frac{1}{6}(14-8 \sqrt{3}-14)^2=32

 

Posted by

sudhir.kumar

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