Question

We have to prove that

Prove that $\frac{\cos A}{1-\tan A}-\frac{{\sin }^{2}A}{\cos A-\sin A}=\sin A+\cos A.$

Answer 1

Prove the given expression.$\frac{\cos A}{1-\tan A}-\frac{{\sin }^{2}A}{\cos A-\sin A}=\sin A+\cos A$

L.H.S. $=\frac{\cos A}{1-\tan A}-\frac{{\sin }^{2}A}{\cos A-\sin A}$

$$\begin{gathered} =\frac{\cos A}{1-{\displaystyle \frac{\sin A}{\cos A}}}-\frac{{\sin }^{2}A}{\cos A-\sin A} \\ =\frac{{\cos }^{2}A}{\cos A-\sin A}-\frac{{\sin }^{2}A}{\cos A-\sin A} \\ =\frac{{\cos }^{2}A-{\sin }^{2}A}{\cos A-\sin A} \\ =\frac{(\cos A-\sin A)(\cos A+\sin A)}{(\cos A-\sin A)} \\ =(\cos A+\sin A) \\ =R.H.S \end{gathered}$$

Hence, L.H.S. = R.H.S. proved.