If tan (A + B) = √3 and tan (A – B) = 1/(√3) ; 0° < A + B < 90°; A > B

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Q) If tan (A + B) = √3 and tan (A – B) = $\frac{1}{\sqrt3}$ ; 0° < A + B < 90°; A > B, find A and B.

Ans:

Given that, tan (A + B) = √3 = tan 60°

Hence, A + B = 60° ………… (i)

Next, its given that, tan (A – B) = $\frac{1}{\sqrt3}$ = tan 30°

Hence, A – B = 30° ………… (i)

By solving equations (i) and (ii), we get,

A = 45⁰and B = 15⁰

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