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प्रश्न
Prove that:
2 sin2 `pi/6` + cosec2 `(7pi)/6` cos2 `pi/3 = 3/2`
उत्तर
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
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