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  • If tan A plus B is equal to sqrt 3 and tan A minus B is equal to 1 sqrt 3 0 degree greater than A plus B leq 90 degree A is lesser than B find A and B

3.     If  \tan (A+B)=\sqrt{3} and \tan (A-B)= \frac{1}{\sqrt{3}};  0^{o}<A+B\leq 90^{o};A> B, find A \: and \: B.

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best_answer

Given that,
\tan (A+B) = \sqrt{3} =\tan 60^0
So, A + B = 60^o..........(i)
\tan (A-B) = 1/\sqrt{3} =\tan 30^0
therefore, A - B = 30^o.......(ii)
By solving the equation (i) and (ii) we get;

A = 45^o and B = 15^o
 

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manish

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