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Question:

Prove that: 2sin²π/6 + cosec²7π/6 cos²π/3 = 3/2

Solution:

LHS=2sin²π/6 + cosec²7π/6 cos²π/3=2sin²π/6 + cosec²(π+π/6)cos²π/3=2sin²π/6 + cosec²(-π/6)cos²π/3As θ lies in the 3rd quadrant, θ is negative.=2(1/2)² + (-2)² × (1/2)²=2 × 1/4 + 1 = 3/2=RHS
Hence proved.