Question
Prove the following.
$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta}$
$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta}$
✓
Answer
Taking LHS
$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}$
Dividing numerator and denominator by cosθ
$=\frac{\frac{\sin \theta}{\cos \theta}-\frac{\cos \theta}{\cos \theta}+\frac{1}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\cos \theta}-\frac{1}{\cos \theta}}$
$=\frac{\tan \theta-1+\sec \theta}{\tan \theta+1-\sec \theta} \times \frac{\sec \theta-\tan \theta}{\sec \theta-\tan \theta}$
$=\frac{(\tan \theta-1+\sec \theta)}{(\tan \theta+1-\sec \theta)} \frac{(\sec \theta-\tan \theta)}{(\sec \theta-\tan \theta)}$
$=\frac{\tan \theta \sec \theta-\tan ^2 \theta-\sec \theta+\tan \theta+\sec ^2 \theta-\tan \theta \sec \theta}{(\tan \theta+1-\sec \theta)(\sec \theta-\tan \theta)}$
$=\frac{\tan \theta+1-\sec \theta+\sec ^2 \theta-\tan ^2 \theta}{(\tan \theta+1-\sec \theta)(\sec \theta-\tan \theta)}$
$=\frac{\tan \theta+1-\sec \theta}{(\tan \theta+1-\sec \theta)(\sec \theta-\tan \theta)}\left[\text { As } \sec ^2 \theta-\tan ^2 \theta=1\right]$
$=\frac{1}{\sec \theta-\tan \theta}$
= RHS
Proved.
$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}$
Dividing numerator and denominator by cosθ
$=\frac{\frac{\sin \theta}{\cos \theta}-\frac{\cos \theta}{\cos \theta}+\frac{1}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\cos \theta}-\frac{1}{\cos \theta}}$
$=\frac{\tan \theta-1+\sec \theta}{\tan \theta+1-\sec \theta} \times \frac{\sec \theta-\tan \theta}{\sec \theta-\tan \theta}$
$=\frac{(\tan \theta-1+\sec \theta)}{(\tan \theta+1-\sec \theta)} \frac{(\sec \theta-\tan \theta)}{(\sec \theta-\tan \theta)}$
$=\frac{\tan \theta \sec \theta-\tan ^2 \theta-\sec \theta+\tan \theta+\sec ^2 \theta-\tan \theta \sec \theta}{(\tan \theta+1-\sec \theta)(\sec \theta-\tan \theta)}$
$=\frac{\tan \theta+1-\sec \theta+\sec ^2 \theta-\tan ^2 \theta}{(\tan \theta+1-\sec \theta)(\sec \theta-\tan \theta)}$
$=\frac{\tan \theta+1-\sec \theta}{(\tan \theta+1-\sec \theta)(\sec \theta-\tan \theta)}\left[\text { As } \sec ^2 \theta-\tan ^2 \theta=1\right]$
$=\frac{1}{\sec \theta-\tan \theta}$
= RHS
Proved.
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