130
Prove that $\frac{\tan A}{1- \cot A} + \frac{\cot A}{1-\tan A} = 1 + \sec A \operatorname{cosec} A$
Show SolutionHide Solution↓
Getting $\frac{\sin^2 A}{\cos A(\sin A-\cos A)} + \frac{\cos^2 A}{\sin A(\cos A-\sin A)}$
$= \frac{\sin^3 A-\cos^3 A}{\sin A \cos A(\sin A-\cos A)}$
$= \frac{(\sin A-\cos A)(\sin^2 A+\cos^2 A+\sin A \cos A)}{\sin A \cos A(\sin A-\cos A)}$
$= \frac{1+\sin A \cos A}{\sin A \cos A} = \frac{1}{\sin A \cos A} + 1$
$= \operatorname{cosec} A \sec A + 1 = \text{RHS}$
$= \frac{\sin^3 A-\cos^3 A}{\sin A \cos A(\sin A-\cos A)}$
$= \frac{(\sin A-\cos A)(\sin^2 A+\cos^2 A+\sin A \cos A)}{\sin A \cos A(\sin A-\cos A)}$
$= \frac{1+\sin A \cos A}{\sin A \cos A} = \frac{1}{\sin A \cos A} + 1$
$= \operatorname{cosec} A \sec A + 1 = \text{RHS}$