Let S = { x ( - 2 , 2 ) : 9 ? 1 - tan 2 x + 9 ? tan 2 x = 10 } and = x S tan 2 ( x 3 ) , then 1 6 ( - 14 ) 2 is equal to [2023]

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Solution

Q.
$Let S = {x \in (- \frac{π}{2}, \frac{π}{2}) : 9^{1 - \tan^{2} x} + 9^{\tan^{2} x} = 10}$ and $β = \sum_{x \in S} \tan^{2} (\frac{x}{3}),$ then $\frac{1}{6} {(β - 14)}^{2}$ is equal to [2023]

1 16

2 8

3 64

4 32

Ans.
(4)

$We have, 9^{1 - \tan^{2} x} + 9^{\tan^{2} x} = 10$

Putting $9^{\tan^{2} x} = t$ , we get

$∴$    $\frac{9}{t} + t = 10 \Rightarrow t^{2} - 10 t + 9 = 0$

$\Rightarrow t^{2} - 9 t - t + 9 = 0 \Rightarrow (t - 1) (t - 9) = 0$ $\Rightarrow t = 1, 9$

$∴$ $9^{\tan^{2} x} = 1 or 9^{\tan^{2} x} = 9$

$\Rightarrow \tan^{2} x = 0 or \tan^{2} x = 1$ $\Rightarrow \tan x = 0 or \tan x = + 1$

$∴$ $x = 0, \frac{π}{4}, - \frac{π}{4}$

$∴$ $x \in (\frac{- π}{2}, \frac{π}{2})$ $...(i)$

Now, $β = \sum \tan^{2} (\frac{x}{3}) = \tan^{2} 0 + \tan^{2} \frac{π}{12} + \tan^{2} (- \frac{π}{12})$

   $= 0 + 2 {(\tan 15 °)}^{2}$

$= 2 {(2 - \sqrt{3})}^{2} = 2 (7 - 4 \sqrt{3})$

$β = 14 - 8 \sqrt{3}$ $...(ii)$

$∴$ $\frac{1}{6} {(β - 14)}^{2} = \frac{1}{6} {(14 - 8 \sqrt{3} - 14)}^{2}$    $[from (ii)]$

$= \frac{192}{6} = 32$

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