If sin θ – cos θ = $$\frac{\sqrt3}{2}$$​, then find the positive value of sin θ + cos θ.

Correct option is D

Given:

​$$\sin \theta - \cos \theta = \frac{\sqrt{3}}{2}$$​​

Concept Used:

​$$\sin^2 \theta + \cos^2 \theta = 1$$​​

Solution:

​$$(\sin \theta - \cos \theta)^2 = \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3}{4}$$​

and $$(\sin \theta - \cos \theta)^2 = 1 - 2 \sin \theta \cos \theta$$​

​$$\frac{3}{4} = 1 - 2 \sin \theta \cos \theta$$​​

2$$\sin \theta \cos \theta = 1 - \frac{3}{4} = \frac{1}{4}$$​​

​$$\sin \theta \cos \theta = \frac{1}{8}$$​

​$$(\sin \theta + \cos \theta)^2 = 1 + 2 \sin \theta \cos \theta = 1 + 2 \times \frac{1}{8} = 1 + \frac{1}{4} = \frac{5}{4}$$​​

​$$\sin \theta + \cos \theta = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}$$​​