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Question

Prove that tan2a-tan2b=[cos2b-cos2a]cos2acos2b=(sin2a-sin2b)(cos2a.cos2b)


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Solution

Determine the proof of the given expression tan2a-tan2b=[cos2b-cos2a]cos2acos2b=(sin2a-sin2b)(cos2a.cos2b)

Solve the L.H.S part:

tan2atan2b=(sin2acos2a)(sin2bcos2b)=(cos2bsin2acos2asin2b)cos2acos2b=[cos2b(1cos2a)cos2a(1cos2b)]cos2acos2bsin2a+cos2a=1=[cos2bcos2bcos2acos2a+cos2acos2b]cos2acos2b=[cos2bcos2a]cos2acos2b=[1sin2b1+sin2a]cos2acos2b=(sin2asin2b)cos2acos2b

Hence, the L.H.S = R.H.S.


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