In right angle ∆ABC, right angled at B, if tanA = $$\sqrt3$$​ then sinAcosC + cosAsinC = ?

Correct option is A

Given:

Right-angled triangle △ABC, right-angled at B

​$$\tan A = \sqrt{3}$$​​

Required: $$\sin A \cos C + \cos A \sin C$$​​

Solution:

​$$\tan A = \frac{\text{Perpendicular }}{\text{Base}} = \sqrt{3}$$​

​$$\frac{\text{BC}}{\text{AB}} = \sqrt{3}$$​​
So, $$∠A = 60^\circ, ∠C= 30^\circ$$

​

Also,

​$$\sin A \cos C + \cos A \sin C = \sin(A + C)$$​

Since $$\angle B = 90^\circ, \angle A +\angle C = 90^\circ$$​​

​$$\sin A \cos C + \cos A \sin C = \sin(A + C) = \sin(90^\circ) = 1$$