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प्रश्न
Evaluate 2 sec2 θ + 3 cosec2 θ – 2 sin θ cos θ if θ = 45°.
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उत्तर
2 sec2 θ + 3 cosec2 θ – 2 sin θ cos θ
= `2(sqrt(2))^2 + 3(sqrt(2))^2 - 2. 1/sqrt(2). 1/sqrt(2)` ...[∵ θ = 45°]
= `2 xx 2 + 3 xx 2 - 2 xx 1/2`
= 4 + 6 – 1
= 9
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