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प्रश्न
Without using tables evaluate: 3cos 80°. cosec 10° + 2sin 59° sec 31°
उत्तर १
3 cos 80°. cosec 10° + 2 sin 59° sec 31°
= 3cos (90° – 10°) cosec 10° + 2sin (90° – 31°) sec 31°
= 3 sin 10° cosec 10° + 2 cos 31°, sec 31° ....[sin (90° – θ) = cos θ, cos (90° – θ) = sin θ]
= 3 × 1 + 2 × 1 .....[∵ sin θ. Cosec θ = 1, cos θ. sec θ = 1]
= 5
उत्तर २
3 cos 80°. cosec 10° + 2 sin 59° sec 31°
⇒ `3 cos (90° - 10°). 1/("sin" 10°) + 2 sin (90° - 31°). 1/(cos 31°)`
⇒ `(3 "sin" 10°)/("sin" 10°) + (2"cos" 31°)/("cos" 31°)`
⇒ 3 + 2 = 5.
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