Let [S=x (-()/(2), ()/(2)): 9^1- ^2 x+9^ ^2 x=10] and [=_x S ^2((x)/(3))], then [(1)/(6)(-…

Question

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

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

Accepted Answer

We have, $$S=\left\{x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right): 9^{1-\tan ^{2} x}+9^{\tan ^{2} x}=10\right\}$$ and

$$\beta=\sum_\limits{x \in S} \tan ^{2}\left(\frac{x}{3}\right)$$

$$
\begin{aligned}
& 9^{1-\tan ^2 x}+9^{\tan ^2 x}=10 \\
&\Rightarrow \frac{9}{9^{\tan ^2 x}}+9^{\tan ^2 x}=10
\end{aligned}
$$

Let $9^{\tan ^2 x}=t$

$$
\frac{9}{t}+t=10 \Rightarrow t^2-10 t+9=0 \Rightarrow t=9 \text { or } 1
$$

If $t=9,9^{\tan ^2 x}=9 \Rightarrow x= \pm \frac{\pi}{4}$

If $t=1,9^{\tan ^2 x}=1 \Rightarrow x=0$

Also,

$$
\begin{aligned}
\beta & =\sum_{x \in S} \tan ^2\left(\frac{x}{3}\right) \\
& =\tan ^2\left(\frac{0}{3}\right)+\tan ^2\left(\frac{\pi}{12}\right)+\tan ^2\left(-\frac{\pi}{12}\right) \\
& =0+2 \tan ^2 \frac{\pi}{12}=2[2-\sqrt{3}]^2 \\
& =2[4+3-4 \sqrt{3}]=14-8 \sqrt{3}
\end{aligned}
$$

$$
\text { So, now } \frac{1}{6}(\beta-14)^2=\frac{1}{6}[14-8 \sqrt{3}-14]^2=\frac{1}{6} \times 64 \times 3=32
$$

Solutions · 1