Prove the identity $\frac{\cos B}{1-\tan B} + \frac{\sin B}{1-\cot B}=\sin B+\cos B$ I have worked on this identity from both sides of the equation and can't seem to get it to equal the other side no matter what I try.
$\displaystyle\frac{\cos B}{1-\tan B} + \frac{\sin B}{1-\cot B} =\sin B+\cos B$

What I would do here is to first convert $\tan B$ and $\cot B$ into $\frac{\sin B}{\cos B}$ and $\frac{\cos B}{\sin B}$, respectively.
$$\frac{\cos B}{1-\tan B}+\frac{\sin B}{1-\cot B}$$
$$=\frac{\cos B}{\left(1-\dfrac{\sin B}{\cos B}\right)}+\frac{\sin B}{\left(1-\dfrac{\cos B}{\sin B}\right)}$$
Simplify the denominators of both fractions.
$$\frac{\cos B}{\left(\dfrac{\cos B}{\cos B}-\dfrac{\sin B}{\cos B}\right)}+\frac{\sin B}{\left(\dfrac{\sin B}{\sin B}-\dfrac{\cos B}{\sin B}\right)}$$
$$=\frac{\cos B}{\left(\dfrac{\cos B - \sin B}{\cos B}\right)}+\frac{\sin B}{\left(\dfrac{\sin B -\cos B}{\sin B}\right)}$$
$$=\frac{\cos^2 B}{\cos B-\sin B}+\frac{\sin^2 B}{\sin B -\cos B}$$
$$=\frac{\cos^2 B}{\cos B-\sin B}+\frac{\sin^2 B}{-\left(\cos B -\sin B\right)}$$
$$=\frac{\cos^2 B}{\cos B-\sin B}-\frac{\sin^2 B}{\cos B-\sin B}$$
Add the two fractions
$$\frac{\cos^2 B-\sin^2 B}{\cos B-\sin B}$$
Factor the numerator using the difference of squares formula (which is $a^2-b^2=(a+b)(a-b)$)
$$\frac{(\cos B+\sin B)(\cos B-\sin B)}{\cos B-\sin B}$$
$$=\cos B+\sin B$$
$$=\sin B+\cos B \ \checkmark$$
$$\color{green}{\therefore \, \, \frac{\cos B}{1-\tan B}+\frac{\sin B}{1-\cot B}=\sin B+\cos B}$$
Hope I helped!
