Question

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

• A. $0$
• B. $1$
• C. $\cos \theta -\sin \theta$
• D. $\cos \theta +\sin \theta$

Answer 1

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

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

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

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

$\Rightarrow$ $\frac{\sin \theta \tan \theta -\cos \theta }{\tan \theta -1}$

$\Rightarrow$ $\frac{\frac{{\sin }^2\theta }{\cos \theta }-\cos \theta }{\tan \theta -1}$

$\Rightarrow$ $\frac{{\sin }^2\theta -{\cos }^2\theta }{\cos \theta (\tan \theta -1)}$

$\Rightarrow$ $\frac{(\sin \theta +\cos \theta )(\sin \theta -\cos \theta )}{\cos \theta (\frac{\sin \theta }{\cos \theta }-1)}$

$\Rightarrow$ $\frac{(\sin \theta +\cos \theta )(\sin \theta -\cos \theta )}{(\sin \theta -\cos \theta )}$

$\Rightarrow$ $\sin \theta +\cos \theta$