Question

$\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}$ is equal to

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

Answer 1

The correct option is B $(\sin A+\cos A)$

Given expression is $\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}$
Multiply first term by $\frac{\cos A}{\cos A}$ and second term by $\frac{\sin A}{\sin A}$,
$=(\frac{\cos A}{1-\tan A}\times \frac{\cos A}{\cos A})+(\frac{\sin A}{1-\cot A}\times \frac{\sin A}{\sin A})$
$=\frac{{\cos }^{2}A}{\cos A-(\frac{\sin A}{\cos A}\times \cos A)}+\frac{{\sin }^{2}A}{\sin A-(\frac{\cos A}{\sin A}\times \sin A)}$
(Since, $\tan A=\frac{\sin A}{\cos A},\cot =\frac{\cos A}{\sin A})$
$=\frac{{\cos }^{2}A}{\cos A-\sin A}+\frac{{\sin }^{2}A}{\sin A-\cos A}$
$=\frac{{\cos }^{2}A}{\cos A-\sin A}-\frac{{\sin }^{2}A}{\cos A-\sin A}$
$=\frac{{\cos }^{2}A-{\sin }^{2}A}{\cos A-\sin A}$
$=\frac{(\cos A-\sin A)(\cos A+\sin A)}{(\cos A-\sin A)}$
$=\cos A+\sin A$
Hence, $\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}=\cos A+\sin A$