$\frac{\sin \theta}{1-\cot \theta}+\frac{\cos \theta}{1-\tan \theta}$ is equal to
(a) 0
(b) 1
(c) $\sin \theta+\cos \theta$
(d) $\sin \theta-\cos \theta$
The given expression is $\frac{\sin \theta}{1-\cot \theta}+\frac{\cos \theta}{1-\tan \theta}$.
Simplifying the given expression, we have
$\frac{\sin \theta}{1-\cot \theta}+\frac{\cos \theta}{1-\tan \theta}$
$=\frac{\sin \theta}{1-\frac{\cos \theta}{\sin \theta}}+\frac{\cos \theta}{1-\frac{\sin \theta}{\cos \theta}}$
$=\frac{\sin \theta}{\frac{\sin \theta-\cos \theta}{\sin \theta}}+\frac{\cos \theta}{\frac{\cos \theta-\sin \theta}{\cos \theta}}$
$=\frac{\sin ^{2} \theta}{\sin \theta-\cos \theta}+\frac{\cos ^{2} \theta}{\cos \theta-\sin \theta}$
$=\frac{\sin ^{2} \theta}{\sin \theta-\cos \theta}+\frac{\cos ^{2} \theta}{-1(\sin \theta-\cos \theta)}$
$=\frac{\sin ^{2} \theta}{\sin \theta-\cos \theta}-\frac{\cos ^{2} \theta}{\sin \theta-\cos \theta}$
$=\frac{\sin ^{2} \theta-\cos ^{2} \theta}{\sin \theta-\cos \theta}$
$=\frac{(\sin \theta+\cos \theta)(\sin \theta-\cos \theta)}{(\sin \theta-\cos \theta)}$
$=\sin \theta+\cos \theta$
Therefore, the correct option is (c).
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.