Question

Let S = {\( x∈(−\frac{ π}{ 2} , \frac{π}{2} )\) : 9^1−tan2 x + 9^tan2 x=10 } and \(β=∑_{ x∈s} t a n ^2 (\frac{ x}{ 3} )\) , then \(\frac{1}{6}\) ( β − 14 )²  is equal to

• A. 8
• B. 16
• C. 32
• D. 34

Answer 1

The Correct Option is C

Step 1: Let \( 9^{\tan^2 x} = P \), so we have the equation: \[ \frac{9}{P} + P = 10 \] Solving for \( P \): \[ P^2 - 10P + 9 = 0 \] \[ (P - 9)(P - 1) = 0 \] Thus, \( P = 9 \) or \( P = 1 \). 
Step 2: Therefore, \( 9^{\tan^2 x} = 9 \), which implies that \( \tan^2 x = 1 \), so \( x = 0, \pm \frac{\pi}{4} \). Thus, \( x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \). 
Step 3: Now, compute \( \beta \): \[ \beta = \tan^2(0) + \tan^2 \left( \frac{\pi}{12} \right) + \tan^2 \left( -\frac{\pi}{12} \right) \] \[ \beta = 0 + 2 \left( \tan 15^\circ \right)^2 \] Using the approximation \( \tan 15^\circ = 2 - \sqrt{3} \), we get: \[ \beta = 2(2 - \sqrt{3})^2 \] \[ \beta = 2(7 - 4\sqrt{3}) \] Now calculate \( \left( \beta - 14 \right)^2 \): \[ \left( \beta - 14 \right)^2 = \left( 14 - 8\sqrt{3} - 14 \right)^2 = 32 \]