JEE Mains · Maths · STD 11 - Trigonometrical equations
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 \(\beta=\sum_{x \in S} \tan ^2\left(\frac{x}{3}\right)\), then \(\frac{1}{6}(\beta-14)^2\) is equal to
- A \(32\)
- B \(8\)
- C \(64\)
- D \(16\)
Answer & Solution
Correct Answer
(A) \(32\)
Step-by-step Solution
Detailed explanation
Let \(9^{\tan ^2 x}= P\) \(\frac{9}{ P }+ P =10\) \(P^2-10 P+9=0\) \((P-9)(P-1)=0\) \(P=1,9\) \(9^{\tan ^2 x}=1,9^{\tan ^2 x}=9\) \(\tan ^2 x =0, \tan ^2 x =1\) \(x =0, \pm \frac{\pi}{4} \quad \therefore x \in\left(-\frac{\pi}{2}, \frac{ p }{2}\right)\)…
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