Question
Prove the following trigonometric identities.
$\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\frac{1}{\sec\theta-\tan\theta}$
$\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\frac{1}{\sec\theta-\tan\theta}$
✓
Answer
Divide Nr and Dr with $\cos\theta,$ we get
$\text{L.H.S}=\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}$
$\frac{\frac{\sin\theta-\cos\theta+1}{\cos\theta}}{\frac{\sin\theta\cos\theta-1}{\cos\theta}}=\frac{\tan\theta-1+\sec\theta}{\tan\theta+1-\sec\theta}$
$=\frac{1}{\frac{\sec\theta-\tan\theta}{\tan\theta-\sec\theta+1}}-1$
$=\frac{1-\sec\theta+\tan\theta}{1-\sec\theta+\tan\theta}\times\frac{1}{\sec\theta-\tan\theta}$
$=\frac{1}{\sec\theta-\tan\theta}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
$\text{L.H.S}=\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}$
$\frac{\frac{\sin\theta-\cos\theta+1}{\cos\theta}}{\frac{\sin\theta\cos\theta-1}{\cos\theta}}=\frac{\tan\theta-1+\sec\theta}{\tan\theta+1-\sec\theta}$
$=\frac{1}{\frac{\sec\theta-\tan\theta}{\tan\theta-\sec\theta+1}}-1$
$=\frac{1-\sec\theta+\tan\theta}{1-\sec\theta+\tan\theta}\times\frac{1}{\sec\theta-\tan\theta}$
$=\frac{1}{\sec\theta-\tan\theta}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
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