What is the value of [(cos 3θ + 2cos 5θ + cos 7θ) ÷ (cos θ + 2cos 3θ + cos 5θ)] + sin 2θ tan 3θ?
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Correct answer: A
Explanation
The correct answer is option1 i.e. cos 2θ
[(cos 3θ + 2cos 5θ + cos 7θ) ÷ (cos θ + 2cos 3θ + cos 5θ)] + sin 2θ tan 3θ
= [(cos 3θ + cos 7θ) +2 cos 5θ] ÷ [(cos θ + cos 5θ) + 2cos 3θ] + sin 2θ tan 3θ
= (2 cos 5θ cos 2θ + 2 cos 5θ)/2cos 3θ cos 2θ + 2 cos 3θ) + sin 2θ tan 3θ
= (2cos 5θ/2cos 3θ) (cos2θ + 1/cos2θ + 1) + sin 2θ tan 3θ
= cos5θ/cos3θ + sin2θ tan3θ/cos3θ
= 2cos5θ - cos5θ + cosθ/2cos3θ = (cos5θ + cosθ)/2cos3θ
= 2cos3θ cos2θ/2cos3θ = cos2θ