Question
Mon April 19, 2010 By:

Please help me with this question

Expert Reply
Mon April 19, 2010

Using the following result,we can write,

tan(C+D) = (tanC + tanD)/(1 - tanC tanD)

tan2A = [tan(A+B) + tan(A-B)]/(1 - tantan(A+B) tan(A-B)) = (x+y)/(1 - xy)

Similarly,

tan2B = [tan(A+B) - tan(A-B)]/(1 + tan(A+B) tan(A-B)) = (x-y)/(1 + xy)

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