Question

cos 2A + 2 cos 4A + cos 6A÷
cos A + 2 cos 3A + cos 5A
= cos A - sin A tan 3A​

Answer 1

LHS - cos 3A + 2 cos 5A + cos 7A / cos A + 2 cos 3A + cos 5A

cos 7A + cos 3A + 2 cos 5A / cos 5A + cos A + 2 cos 3A

(cos 7A + cos 3A) + 2 cos 5A / (cos 5A + cos A) + 2 cos 3A

(2 cos 7A+3A/2 cos 7A-3A/2) + 2 cos 5A / (2 cos 5A+A/2 cos 5A-A/2) + 2 cos 3A

(2 cos 10A/2 cos 4A/2) + 2 cos 5A / (2 cos 6A/2 cos 4A/2) + 2 cos 3A

2 cos 5A cos 2A + 2 cos 5A / 2 cos 3A cos 2A + 2 cos 3A

2 cos 5A (cos 2A + 1) / 2 cos 3A (cos 2A + 1)

Cos 5A / cos 3A

Cos (3A+2A) / cos 3A

Cos 3A cos 2A – Sin 3A Sin 2A / cos 3A

Cos 3A cos 2A / cos 3A – Sin 3A Sin 2A / cos 3A [separating both sides]

Cos 2A – tan 3A Sin 2A [ Sin 3A / cos 3A = tan 3A]

cos 2A – Sin 2A tan 3A

RHS

Hare Krishna

Answer 2

Step-by-step explanation:

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