​$$\text{If } \sin\theta + \cos\theta = \frac{\sqrt{7}}{2}, \text{ then find the value of } \sin\theta - \cos\theta.$$​​

Correct option is C

Given:
​$$\sin \theta + \cos \theta = \frac{\sqrt{7}}{2}$$

Formula Used:

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

Solution:

​$$\text{Step 1: Square both sides of the given equation:}\\(\sin \theta + \cos \theta)^2 = \left(\frac{\sqrt{7}}{2}\right)^2\\\sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta = \frac{7}{4}\\\\\text{Step 2: Substitute } \sin^2 \theta + \cos^2 \theta = 1:\\1 + 2\sin \theta \cos \theta = \frac{7}{4}\\2\sin \theta \cos \theta = \frac{7}{4} - 1 = \frac{3}{4}\\\sin \theta \cos \theta = \frac{3}{8}\\\\\text{Step 3: Use the formula for } \sin \theta - \cos \theta:\\(\sin \theta - \cos \theta)^2 = 1 - 2\sin \theta \cos \theta\\(\sin \theta - \cos \theta)^2 = 1 - 2 \times \frac{3}{8} = 1 - \frac{6}{8} = \frac{2}{8} = \frac{1}{4}\\\sin \theta - \cos \theta = \sqrt{\frac{1}{4}} = \frac{1}{2}\\\text{Thus, the value of } \sin \theta - \cos \theta \text{ is } \frac{1}{2}.$$​​