If , where , then is equal…

MATHEMATICS · Single correct · CBSE · Mathematics · Introduction to Trigonometry · Introduction to Trigonometry

Problem

If ksin⁡2x+1kcsc⁡2x=2k\sin^2x+\frac{1}{k}\csc^2x=2, where x∈(0,π2)x\in\left(0,\frac{\pi}{2}\right), then cos⁡2x+5sin⁡xcos⁡x+sin⁡2x\cos^2x+5\sin x\cos x+\sin^2x is equal to (1) k2+5k+6k2\frac{k^2+5k+6}{k^2} (2) k2−5k+6k2\frac{k^2-5k+6}{k^2} (3) 66 (4) None of these

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