JEE PYQ: Trigonometric Ratios And Identities Question 11
Question 11 - 2019 (09 Jan Shift 1)
For any $\theta \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$, the expression $3(\sin\theta - \cos\theta)^4 + 6(\sin\theta + \cos\theta)^2 + 4\sin^6\theta$ equals:
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(1) $13 - 4\cos^2\theta + 6\sin^2\theta\cos^2\theta$
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(2) $13 - 4\cos^6\theta$
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(3) $13 - 4\cos^6\theta + 6\cos^4\theta$
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(4) $13 - 4\cos^4\theta + 2\sin^2\theta\cos^2\theta$
Show Answer
Answer: (2) $13 - 4\cos^6\theta$
Solution
Expanding: $= 9 + 12\sin^2\theta\cos^2\theta + 4\sin^6\theta = 9 + 4 - 4\cos^6\theta = 13 - 4\cos^6\theta$.
BETA
