Question

If tan (A + B) = √3, tan (A − B) = (1/√3), 0° < A + B < 90°, A > B, find A and B. Also calculate, tanA sin(A + B) + cos A tan (A − B).

Answer 1

Explanation:

sorry I am understanding

Answer 2

A = 45° and B = 15°

Explanation:

Given:  

tan (A + B) = √3

tan (A − B) = 1/√3

0° < A + B < 90°, A > B

We know that tan 60° = √3.

So tan (A + B) = √3 = tan 60°  

A + B = 60° ------(1)

We know that tan 30° = 1/√3

So tan (A - B) = 1/√3 = tan 30°  

A - B = 30° ------(2)

Adding equations 1 & 2, we get: 2A = 90°

A = 90° / 2 = 45°  

Substituting A in equation 1, we get:  

45° + B = 60°

B = 60° -  45° = 15°

Solved. A = 45° and B = 15° satisfies all conditions 0° < A + B < 90°, A > B.

tanA sin(A + B) + cos A tan (A − B) = tan45° sin(45°+ 15°) + cos 45° tan (45° − 15°)

= tan45° * sin60° + cos 45° * tan 30°

= (1 * √3 / 2 ) + (1 / √2 * 1/√3 )

= √3 / 2 + 1/√2.√3

= √2√3 + 1 / 2√3