RHS= 3/2
LHS= 2sin²π/6+cosec²7π/6cos²π/3
= 2sin²π/6+cosec² (π+π/6) cos²π/3
= 2sin²π/6+cosec²(-π/6)cos²π/3
As θ lies in the 3rd quadrant, θ is negative.
=2 (½)²+(-2)²× (½)²
2×(¼)+1 = 3/2
2sin²π/6+cosec²7π/6cos²π/3= 3/2
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Verified answer
TO PROVE THAT LHS = RHS
RHS= 3/2
LHS= 2sin²π/6+cosec²7π/6cos²π/3
= 2sin²π/6+cosec² (π+π/6) cos²π/3
= 2sin²π/6+cosec²(-π/6)cos²π/3
As θ lies in the 3rd quadrant, θ is negative.
=2 (½)²+(-2)²× (½)²
2×(¼)+1 = 3/2
hence proved that,
2sin²π/6+cosec²7π/6cos²π/3= 3/2
HOPE IT HELPS YOU