Question

The expression $\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}$ can be written as :

• A. $\sin A\cdot \cos A+1$
• B. $\sec A\cdot \text{cosec}A+1$
• C. $\tan A+\cot A$
• D. $\sec A+\text{cosec}A$

Answer 1

The correct option is B $\sec A\cdot \text{cosec}A+1$

$\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}$
$=\frac{{\sin }^{2}A}{\cos A(\sin A-\cos A)}+\frac{{\cos }^{2}A}{\sin A(\cos A-\sin A)}$
$=\frac{1}{\sin A-\cos A}[\frac{{\sin }^{2}A}{\cos A}-\frac{{\cos }^{2}A}{\sin A}]$
$=\frac{1}{\sin A-\cos A}\cdot \frac{{\sin }^{3}A-{\cos }^{3}A}{\sin A\cos A}$
$=\frac{(\sin A-\cos A)({\sin }^{2}A+{\cos }^{2}A+\sin A\cos A)}{(\sin A-\cos A)\sin A\cos A}$
$=\frac{1+\sin A\cos A}{\sin A\cos A}$
$=\sec A\cdot \text{cosec}A+1$