Question

Cos A minus Cos 3A / sin 3A minus sin Ais equal to tan 2 A

Answer 1

●TRIGONOMETRIC IDENTITY :-

$= > coa \alpha - cos \beta = - 2sin( \frac{ \alpha + \beta }{2} ).sin (\frac{ \alpha - \beta }{2} ) \\ \\ AND \\ \\ = > sin \alpha - sin \beta = 2cos( \frac{ \alpha + \beta }{2} ).sin( \frac{ \alpha - \beta }{2} )$
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●YOUR QUESTION IS :-

$= > \frac{cosA - cos3A}{sin3A - sinA} = tan2A$
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●WE TAKE LEFT HAND SIDE :-

$= > \frac{cosA - cos3A}{sin3A - sinA} \\ \\ = > \frac{ - 2sin \frac{(A + 3A)}{2} .sin \frac{(A - 3A)}{2} }{2cos \frac{(3A + A)}{2}.sin \frac{(3A - A)}{2} } \\ \\ = > \frac{ - 2sin( \frac{4A}{2}).sin( \frac{ - 2A}{2}) }{2cos( \frac{4A}{2}).sin( \frac{2A}{2} )} \\ \\ = > \frac{2sin2A.sinA}{2cos2A.sinA} \\ \\ = > tan2A$
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______☆$BE \: \: \: BRAINLY$☆______

Answer 2

Heya mate
here is your answer

I hope this will help you...✌️✌️
thank you:-)