Question
Prove that:
$\frac{\tan \theta}{\sec \theta-1}=\frac{\tan \theta+\sec \theta+1}{\tan \theta+\sec \theta-1}$
$\frac{\tan \theta}{\sec \theta-1}=\frac{\tan \theta+\sec \theta+1}{\tan \theta+\sec \theta-1}$
✓
Answer
Taking RHS
$=\left(\frac{\tan \theta+\sec \theta+1}{\tan \theta+\sec \theta-1}\right) \times\left(\frac{\sec \theta-1}{\sec \theta-1}\right)$
$=\frac{\tan \theta \sec \theta-\tan \theta+\sec ^2 \theta-\sec \theta+\sec \theta-1}{(\tan \theta+\sec \theta-1)(\sec \theta-1)}$
$=\frac{\tan \theta \sec \theta-\tan \theta+\sec ^2 \theta-1}{(\tan \theta+\sec \theta-1)(\sec \theta-1)}$
$=\frac{\tan \theta \sec \theta-\tan \theta+\tan ^2 \theta}{(\tan \theta+\sec \theta-1)(\sec \theta-1)}\left[ As \sec ^2 \theta-1=\tan ^2 \theta\right]$
$=\frac{\tan \theta(\tan \theta+\sec \theta-1)}{(\tan \theta+\sec \theta-1)(\sec \theta-1)}$
$=\frac{\tan \theta}{(\sec \theta-1)}$
$=$ LHS
Proved.
$=\left(\frac{\tan \theta+\sec \theta+1}{\tan \theta+\sec \theta-1}\right) \times\left(\frac{\sec \theta-1}{\sec \theta-1}\right)$
$=\frac{\tan \theta \sec \theta-\tan \theta+\sec ^2 \theta-\sec \theta+\sec \theta-1}{(\tan \theta+\sec \theta-1)(\sec \theta-1)}$
$=\frac{\tan \theta \sec \theta-\tan \theta+\sec ^2 \theta-1}{(\tan \theta+\sec \theta-1)(\sec \theta-1)}$
$=\frac{\tan \theta \sec \theta-\tan \theta+\tan ^2 \theta}{(\tan \theta+\sec \theta-1)(\sec \theta-1)}\left[ As \sec ^2 \theta-1=\tan ^2 \theta\right]$
$=\frac{\tan \theta(\tan \theta+\sec \theta-1)}{(\tan \theta+\sec \theta-1)(\sec \theta-1)}$
$=\frac{\tan \theta}{(\sec \theta-1)}$
$=$ LHS
Proved.
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