Question

Prove that $\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}-\sin A-\cos A=0$

Answer 1

Given,

$\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}=\sin A+\cos A$ ........(1)

$\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}=\frac{{\cos }^{2}A}{\cos A-\sin A}+\frac{{\sin }^{2}A}{\sin A-\cos A}$

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

$\frac{{\cos }^{2}A-{\sin }^{2}A}{\cos A-\sin A}=\frac{(\cos A+\sin A)(\cos A-\sin A)}{(\cos A-\sin A)}=\cos A+\sin A$.......(2)

Equating (1) and (2)

$\sin A+\cos A=\sin A+\cos A$

$=0$

Hence,

$\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}-\sin A-\cos A=0$