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प्रश्न
Evaluate:
sin2 34° + sin2 56° + 2 tan 18° tan 72° – cot2 30°
उत्तर
sin2 34° + sin2 56° + 2 tan 18° tan 72° – cot2 30°
= sin2 34° + [sin2 (90° – 34°)]2 + 2 tan 18° tan (90° – 18°) – cot2 30°
= `sin^2 34^circ + cos^2 34^circ + 2 tan 18^circ cot 18^circ - (sqrt(3))^2`
= `1 + 2 tan 18^circ xx 1/(tan 18^circ) - 3` ...[∵ sin2θ + cos2θ = 1]
= 1 + 2 – 3
= 3 – 3
= 0
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