Question

Prove that:

2 sin² `pi/6` + cosec² `(7pi)/6`  cos² `pi/3 = 3/2`

Answer 1

LHS = 2 sin² `pi/6` + cosec² `(7pi)/6`  cos² `pi/3`

[∵ `(7pi)/6` = 210°, 210° = 180° + 30°. For 180° + 30° no change in T-ratio. 210° lies in 3rd quadrant, cosec θ is negative.]

`= 2(sin pi/6)^2 + ("cosec" (180^circ + 30^circ))^2 (cos pi/3)^2`

`= 2(1/2)^2 + (- "cosec"  30^circ)^2 * (1/2)^2`

`= 2 xx 1/4 + (- 2)^2 1/4`

`= 2/4 + 4/4 = 6/4`

`= 6/4`

`= 3/2`

= RHS