If tan (A + B) = \(\sqrt3\) and tan (A - B) = \(\frac{1}{\sqrt3}\); 0°\( <\) (A + B) \(≤\) 90° ; A > B, find A and B

Question

Evaluate the following.
(i) sin60° cos30° + sin30° cos 60°
(ii) 2tan²45° + cos²30° - sin²60°
(iii) \(\frac{cos 45°}{sec 30°+cosec30°}\)

(iv) \(\frac{sin\ 30°+tan\ 45°cosec\ 60°}{sec\ 30°+cos\ 60°+cot\ 45°}\)

(v) \(\frac{5cos^260°+4sec^230°-tan^245°}{sin^230°+cos^230°}\)

Answer 1

Given tan(A + B) = \(\sqrt3\), which implies tan(A + B) = 60°. Thus, A + B = 60° (1).

Given tan(A - B) = \(\frac{1}{\sqrt3}\), which implies tan(A - B) = tan 30°. Thus, A - B = 30° (2).

Adding equations (1) and (2): 2A = 90°, so A = 45°.

Substituting A = 45° into equation (1): 45° + B = 60°, so B = 15°.

Therefore, ∠A = 45° and ∠B = 15°.

Solution and Explanation