JEE Main 2023 — Mathematics Trigonometry

Given,

$${9}^{1-{\mathrm{tan}}^{2}x}+{9}^{{\mathrm{tan}}^{2}x}=10$$91−tan2x+9tan2x=10

Now let ${9}^{{\mathrm{tan}}^{2}x}=t$9tan2x=t, then above equation will be,

$$\frac{9}{t}+t=10$$t9​+t=10

$$\Rightarrow {t}^{2}-10t+9=0$$⇒t2−10t+9=0

$$\Rightarrow t=9\text{or}t=1$$⇒t=9ort=1

So, when ${9}^{{\mathrm{tan}}^{2}x}=9\Rightarrow {\mathrm{tan}}^{2}x=1$9tan2x=9⇒tan2x=1

$\Rightarrow \mathrm{tan}x=\pm 1\Rightarrow x=\pm \frac{\pi }{4}$⇒tanx=±1⇒x=±4π​, as given $x\in (\frac{-\pi }{2},\frac{\pi }{2})$x∈(2−π​,2π​)

Now when ${9}^{{\mathrm{tan}}^{2}x}=1$9tan2x=1

$$\Rightarrow {\mathrm{tan}}^{2}x=0\Rightarrow x=0$$⇒tan2x=0⇒x=0

Hence, $\displaystyle \beta =\underset{x\in S}{\sum }(\frac{x}{3})={\mathrm{tan}}^{2}(\frac{0}{3})+{\mathrm{tan}}^{2}(\frac{\pi }{12})+{\mathrm{tan}}^{2}(\frac{-\pi }{12})$β=x∈S∑​(3x​)=tan2(30​)+tan2(12π​)+tan2(12−π​)

$$=0+2{(2-\sqrt{3})}^{2}$$=0+2(2−3​)2

$$=14-8\sqrt{3}$$=14−83​

So, the value of $\frac{1}{6}{(\beta -14)}^{2}=\frac{64\times 3}{6}=32$61​(β−14)2=664×3​=32

Hence this is the correct option.