​$$\frac{\cos\theta}{1 - \tan\theta} + \frac{\sin\theta}{1 - \cot\theta}$$ = ?​

Correct option is D

Given:

​$$\frac{\cos\theta}{1 - \tan\theta} + \frac{\sin\theta}{1 - \cot\theta}$$​​

Formula Used:

​$$\tan\theta = \frac{\sin\theta}{\cos\theta}, \quad \cot\theta = \frac{\cos\theta}{\sin\theta}$$​​

Solution:

$$\text{First term:}\\ \quad \frac{\cos\theta}{1 - \tan\theta} = \frac{\cos\theta}{1 - \frac{\sin\theta}{\cos\theta}} \\= \frac{\cos\theta}{\frac{\cos\theta - \sin\theta}{\cos\theta}} = \frac{\cos^2\theta}{\cos\theta - \sin\theta}\\\ \\\text{Second term:} \quad\\ \frac{\sin\theta}{1 - \cot\theta} \\= \frac{\sin\theta}{1 - \frac{\cos\theta}{\sin\theta}} = \frac{\sin\theta}{\frac{\sin\theta - \cos\theta}{\sin\theta}} = \frac{\sin^2\theta}{\sin\theta - \cos\theta}\\\ \\\text{Now add both:} \quad \\\frac{\cos^2\theta}{\cos\theta - \sin\theta} + \frac{\sin^2\theta}{\sin\theta - \cos\theta} = \frac{\cos^2\theta - \sin^2\theta}{\cos\theta - \sin\theta}\\\ \\= \frac{(\cos\theta + \sin\theta)(\cos\theta - \sin\theta)}{\cos\theta - \sin\theta} = \cos\theta + \sin\theta$$​​

Final Answer: (d) cosθ + sinθ