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प्रश्न
Evaluate: `sin 18^@/cos 72^@ + sqrt3 [tan 10° tan 30° tan 40° tan 50° tan 80°]`
उत्तर
sin 18° = sin (90° - 72) = cos 72°
tan 10° = cot 80° tan 50° = cot 40°
`=> sin 18^@/sin 18^@ + sqrt3 [tan 80 cos 30 . tan 40 cot 40 . 1/sqrt3]`
`= 1 + sqrt3 . 1/sqrt3 = 2`
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