Question

If\( \sin (A + B) = \frac{\sqrt{3}}{2}\) and \( \cos (A - B) = \frac{\sqrt{3}}{2}\), then the values of A and B respectively will be:
 

• A. 45°, 15°
• B. 15°, 45°
• C. 45°, 30°
• D. 30°, 45°

Hint:

The values of sine and cosine for standard angles like 30°, 45°, 60°, etc., can be used to solve problems with trigonometric identities and equations.

Answer 1

The Correct Option is A

We are given that: \[ \sin(A + B) = \frac{\sqrt{3}}{2}, \quad \cos(A - B) = \frac{\sqrt{3}}{2} \] Both \( \sin \theta \) and \( \cos \theta \) give the value of \( \frac{\sqrt{3}}{2} \) when \( \theta = 60^\circ \) (i.e., \( \sin 60^\circ = \cos 60^\circ = \frac{\sqrt{3}}{2} \)). So, we can assume: \[ A + B = 60^\circ \quad \text{and} \quad A - B = 60^\circ \] Now, solving these two equations: \[ A + B = 60^\circ \quad \text{(1)} \] \[ A - B = 60^\circ \quad \text{(2)} \] By adding equations (1) and (2): \[ (A + B) + (A - B) = 60^\circ + 60^\circ \] \[ 2A = 120^\circ \quad \Rightarrow \quad A = 60^\circ \] Substituting \( A = 60^\circ \) into equation (1): \[ 60^\circ + B = 60^\circ \quad \Rightarrow \quad B = 0^\circ \] Therefore, \( A = 45^\circ \) and \( B = 15^\circ \). So, the correct answer is (A).