Let S=x∈−π2,π2:91−tan2⁡x+9tan2⁡x=10 and β=∑x∈s tan2⁡x3, then 16(β−14)2 is equal to

91−tan2⁡x+9tan2⁡x=10 let 9tan2⁡x=t⇒9t+t=10⇒9+t2=10t⇒t2−10t+9=0⇒(t−1)(t−9)=0t=1 or t=99tan2⁡x=1 or 9tan2⁡x=9⇒tan2⁡x=0 or tan2⁡x=1⇒tan⁡x=0 or tan⁡x=±1x=0,π4,−π4⇒∑tan2⁡x3=tan2⁡0+tan2⁡π12+tan2⁡−π12=0+2tan2⁡π12=2tan2⁡150=2(2−3)2=14−83⇒16(β−14)2=16(−83)2=8×8×36=32

Let S=x∈−π2,π2:91−tan2⁡x+9tan2⁡x=10 and β=∑x∈s tan2⁡x3, then 16(β−14)2 is equal to

91−tan2⁡x+9tan2⁡x=10 let 9tan2⁡x=t⇒9t+t=10⇒9+t2=10t⇒t2−10t+9=0⇒(t−1)(t−9)=0t=1 or t=99tan2⁡x=1 or 9tan2⁡x=9⇒tan2⁡x=0 or tan2⁡x=1⇒tan⁡x=0 or tan⁡x=±1x=0,π4,−π4⇒∑tan2⁡x3=tan2⁡0+tan2⁡π12+tan2⁡−π12=0+2tan2⁡π12=2tan2⁡150=2(2−3)2=14−83⇒16(β−14)2=16(−83)2=8×8×36=32

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$Let S = \{x \in (- \frac{π}{2}, \frac{π}{2}) : 9^{1 - \tan^{2} x} + 9^{\tan^{2} x} = 10\} and β = \sum_{x \in s} \tan^{2} (\frac{x}{3}), then \frac{1}{6} (β - 14)^{2} is$ equal to

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Detailed Solution

$\begin{array}{l} 9^{1 - \tan^{2} x} + 9^{\tan^{2} x} = 10 let 9^{\tan^{2} x} = t \\ \Rightarrow \frac{9}{t} + t = 10 \Rightarrow 9 + t^{2} = 10 t \\ \Rightarrow t^{2} - 10 t + 9 = 0 \Rightarrow (t - 1) (t - 9) = 0 \\ t = 1 or t = 9 \\ 9^{\tan^{2} x} = 1 or 9^{\tan^{2} x} = 9 \\ \Rightarrow \tan^{2} x = 0 or \tan^{2} x = 1 \\ \Rightarrow \tan x = 0 or \tan x = \pm 1 \\ x = 0, \frac{π}{4}, - \frac{π}{4} \\ \Rightarrow \sum \tan^{2} (\frac{x}{3}) = \tan^{2} 0 + \tan^{2} (\frac{π}{12}) + \tan^{2} (\frac{- π}{12}) \\ = 0 + 2 \tan^{2} (\frac{π}{12}) = 2 \tan^{2} 15^{0} = 2 (2 - \sqrt{3})^{2} = 14 - 8 \sqrt{3} \\ \Rightarrow \frac{1}{6} (β - 14)^{2} = \frac{1}{6} (- 8 \sqrt{3})^{2} = \frac{8 \times 8 \times 3}{6} = 32 \end{array}$