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Prove that : $\frac{\sin \theta - \cos \theta + 1}{\cos \theta + \sin \theta - 1} = \frac{1}{\sec \theta - \tan \theta}$
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LHS = $\frac{\sin \theta - \cos \theta + 1}{\cos \theta + \sin \theta - 1}$
Dividing Numerator and Denominator by $\cos \theta$,
$\frac{\tan \theta - 1 + \sec \theta}{1 + \tan \theta - \sec \theta}$
$\frac{(\tan \theta + \sec\theta)-(\sec^2\theta-\tan^2\theta)}{1+\tan \theta - \sec \theta}$
$\frac{(\tan \theta + \sec\theta)-(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)}{1+\tan \theta - \sec \theta}$
$\frac{(\tan \theta + \sec\theta) (1-\sec\theta+\tan\theta)}{1+\tan \theta - \sec \theta}$
$= (\tan \theta + \sec \theta)$
Multiplying & dividing by $(\sec \theta – \tan \theta)$
$= (\tan \theta + \sec \theta) \times \frac{(\sec \theta - \tan \theta)}{(\sec \theta - \tan \theta)}$
$= \frac{(\sec^2\theta-\tan^2\theta)}{\sec \theta - \tan \theta} = \frac{1}{\sec \theta - \tan \theta} = \text{RHS}$
Dividing Numerator and Denominator by $\cos \theta$,
$\frac{\tan \theta - 1 + \sec \theta}{1 + \tan \theta - \sec \theta}$
$\frac{(\tan \theta + \sec\theta)-(\sec^2\theta-\tan^2\theta)}{1+\tan \theta - \sec \theta}$
$\frac{(\tan \theta + \sec\theta)-(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)}{1+\tan \theta - \sec \theta}$
$\frac{(\tan \theta + \sec\theta) (1-\sec\theta+\tan\theta)}{1+\tan \theta - \sec \theta}$
$= (\tan \theta + \sec \theta)$
Multiplying & dividing by $(\sec \theta – \tan \theta)$
$= (\tan \theta + \sec \theta) \times \frac{(\sec \theta - \tan \theta)}{(\sec \theta - \tan \theta)}$
$= \frac{(\sec^2\theta-\tan^2\theta)}{\sec \theta - \tan \theta} = \frac{1}{\sec \theta - \tan \theta} = \text{RHS}$