Question
If tan A = a tan B and sin A = b sin B, then the value
of cos2A is:Solution
ATQ, tan A = a tan B 
Since, c A = b sin B ⇒ sin A/sin B = b 
⇒ b cos B = a cos A on squaring both sides, ⇒ b2 cos2 B = a2 cos2A  ---(ii) Now, we have, sin A = b sin B ⇒ sin2A = b2 sin2 B ⇒ 1 - cos2 A = b2 (1 - cos2B) ⇒ 1 - cos2 A = b 2  - b2 cos 2 B ⇒ b2 cos 2 B = b2 - 1 + cos2 A  From equation (ii), ⇒ b2  cos2 B = a 2  cos 2 A ⇒ b2  - 1 + cos2 A  = a 2  cos 2 A ⇒ b2  - 1 = a 2  cos 2 A -  cos 2 A ⇒ b 2  - 1 = cos2A (a2 -1) 
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