Question

If tan (A − B) = $\frac{1}{\sqrt{3}}$ and tan (A + B) = $\sqrt{3}$, 0° < (A + B) < 90° and A > B, then find A and B.

Answer 1

Here, tan (A − B) = $\frac{1}{\sqrt{3}}$
⇒ tan (A − B) = tan 30^o [∵ tan 30^o = $\frac{1}{\sqrt{3}}$]
⇒ A − B = 30^o ...(i)

Also, tan (A + B) = $\sqrt{3}$
⇒​ tan (A + B) = tan 60^o [∵ tan 60^o = $\sqrt{3}$]
⇒ A + B = 60^o ...(ii)

Solving (i) and (ii), we get:
A = 45^o and B = 15^o