Question

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

• A. $\sin \theta -\cos \theta$
• B. $\sin \theta +\cos \theta$
• C. $\tan \theta +\cot \theta$
• D. $\tan \theta -\cot \theta$

Answer 1

The correct option is B $\sin \theta +\cos \theta$

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

$=\frac{\cos \theta }{1-\frac{\sin \theta }{\cos \theta }}+\frac{\sin \theta }{1-\frac{\cos \theta }{\sin \theta }}$

$=\frac{{\cos }^2\theta }{\cos \theta -\sin \theta }-\frac{{\sin }^2\theta }{\cos \theta -\sin \theta }$

$=\frac{{\cos }^2\theta -{\sin }^2\theta }{\cos \theta -\sin \theta }=\cos \theta +\sin \theta$