If tan(A + B) = √3 and tan(A − B) = 1/√3 ; 0° < A + B < 90° ; A > B, find A and B.

Question

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

Answer 1

Given that tan(A + B) = \(\sqrt{3}\) and tan(A − B) = \(\frac{1}{\sqrt{3}}\) where 0° < A + B < 90° ; A > B.

Since, tan(A + B) = \(\sqrt{3}\) = tan 60° and 0° < A + B < 90°.

Therefore, A + B = 60°. ...(1)

And tan(A - B) = \(\frac{1}{\sqrt{3}}\) = tan 30°. Therefore, A - B = 30°. ...(2)

By adding equations (1) and (2), we get A + B + (A − B) = 60° + 30°

⇒ 2A = 90° ⇒ A = 45°.

Now, putting the value of A = 45° in equation (1), we get

45° + B = 60° ⇒ B = 60° − 45° = 15°.

Hence, A = 45° and B = 15°.