176
Prove that : $\frac{\tan \theta}{1-\cot \theta} + \frac{\cot \theta}{1-\tan \theta} = 1 + \sec \theta \operatorname{cosec} \theta$
Show SolutionHide Solution↓
LHS = $\frac{\tan \theta}{1-\cot \theta} + \frac{\cot \theta}{1-\tan \theta}$
$= \frac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1-\frac{\sin \theta}{\cos \theta}}$ (1/2)
$= \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)}$ (1)
$= \frac{1}{(\sin \theta - \cos \theta)} \left[\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta}\right]$ (1/2)
$= \frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)}{(\sin \theta - \cos \theta) \sin \theta \cos \theta}$ (1/2)
$= \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta}$
$= \frac{1}{\sin \theta \cos \theta} + 1 = 1 + \sec \theta \operatorname{cosec} \theta$ = RHS (1/2)
$= \frac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1-\frac{\sin \theta}{\cos \theta}}$ (1/2)
$= \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)}$ (1)
$= \frac{1}{(\sin \theta - \cos \theta)} \left[\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta}\right]$ (1/2)
$= \frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)}{(\sin \theta - \cos \theta) \sin \theta \cos \theta}$ (1/2)
$= \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta}$
$= \frac{1}{\sin \theta \cos \theta} + 1 = 1 + \sec \theta \operatorname{cosec} \theta$ = RHS (1/2)