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Evaluate: sin2 34° + sin2 56° + 2 tan 18° tan 72° – cot2 30° - Mathematics

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Question

Evaluate:

sin2 34° + sin56° + 2 tan 18° tan 72° – cot30°

Evaluate

Solution

sin2 34° + sin56° + 2 tan 18° tan 72° – cot30° 

= 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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Chapter 21: Trigonometrical Identities - Exercise 21 (C) [Page 329]

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Selina Mathematics [English] Class 10 ICSE
Chapter 21 Trigonometrical Identities
Exercise 21 (C) | Q 11 | Page 329
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