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If sin⁡θ+csc⁡θ=2\sin\theta + \csc\theta = 2, then what is the value of sin⁡2θ+csc⁡2θ\sin^2\theta + \csc^2\theta?

Solution

✅ Correct Option: 3

Squaring both sides: (sin⁡θ+csc⁡θ)2=4(\sin\theta + \csc\theta)^2 = 4, so sin⁡2θ+csc⁡2θ+2sin⁡θ⋅csc⁡θ=4\sin^2\theta + \csc^2\theta + 2\sin\theta \cdot \csc\theta = 4.

Since sin⁡θ⋅csc⁡θ=1\sin\theta \cdot \csc\theta = 1: sin⁡2θ+csc⁡2θ=4−2=2\sin^2\theta + \csc^2\theta = 4 - 2 = 2.

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