Home › MCQs › EEJ MAIN Mathematics › Question #7469
Back to Questions
EEJ MAIN Mathematics QUESTION #7469
Question 1
For any \(\theta \in (\frac{\pi}{4}, \frac{\pi}{2})\), the expression \(3(\sin\theta - \cos\theta)^4 + 6(\sin\theta + \cos\theta)^2 + 4\sin^6\theta\) equals:
  • 13 - 4cos²θ + 6sin²θcos²θ✔️
  • 13 - 4cos⁶θ
  • 13 - 4cos²θ + 6cos⁴θ
  • 13 - 4cos⁴θ + 2sin²θcos²θ
Correct Answer Explanation
Expanding each term: \((\sin\theta-\cos\theta)^4 = \sin^4\theta - 4\sin^3\theta\cos\theta + 6\sin^2\theta\cos^2\theta - 4\sin\theta\cos^3\theta + \cos^4\theta\), \((\sin\theta+\cos\theta)^2 = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = 1 + 2\sin\theta\cos\theta\). Combining and simplifying using \(\sin^2\theta + \cos^2\theta = 1\) and algebraic manipulation gives: \(13 - 4\cos^2\theta + 6\sin^2\theta\cos^2\theta\).