Question

\(\frac{\cos\theta}{1-\tan\theta} + \frac{\sin\theta}{1-\cot\theta}\) is equal to

• A. \(\sec\theta + \csc\theta\)
• B. \(\sin\theta + \cos\theta\)
• C. \(\tan\theta + \cot\theta\)
• D. \(\sin\theta - \cos\theta\)

Hint:

Convert tan and cot into sin and cos to simplify expressions easily.

Answer 1

The Correct Option is B

Concept: Convert \(\tan\theta\) and \(\cot\theta\) into \(\sin\theta, \cos\theta\).

Step 1: Rewrite.
\[ \frac{\cos\theta}{1-\frac{\sin\theta}{\cos\theta}} + \frac{\sin\theta}{1-\frac{\cos\theta}{\sin\theta}} \]

Step 2: Simplify.
\[ = \frac{\cos^2\theta}{\cos\theta-\sin\theta} + \frac{\sin^2\theta}{\sin\theta-\cos\theta} \] \[ = \frac{\cos^2\theta}{\cos\theta-\sin\theta} - \frac{\sin^2\theta}{\cos\theta-\sin\theta} \] \[ = \frac{\cos^2\theta - \sin^2\theta}{\cos\theta-\sin\theta} \] \[ = \frac{(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)}{\cos\theta-\sin\theta} \] \[ = \sin\theta + \cos\theta \]