Question

Prove that : \(\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta}\)

Hint:

An alternative and often easier way to solve this identity is to work backwards from the Right-Hand Side (RHS):
\[ \text{RHS} = \frac{1}{\sec \theta - \tan \theta} = \sec \theta + \tan \theta = \frac{1 + \sin \theta}{\cos \theta} \] Now, multiply both the numerator and denominator by \((\sin \theta + \cos \theta - 1)\) and simplify.
This algebraic expansion directly produces the Left-Hand Side (LHS)!

Answer 1

Step 1: Convert the claim into a product statement.
Proving \(\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1} = \frac{1}{\sec\theta-\tan\theta}\) is the same as proving: \[ (\sin\theta - \cos\theta + 1)(\sec\theta - \tan\theta) = \sin\theta + \cos\theta - 1 \]
Step 2: Expand the left side term by term.
\[ \sin\theta\sec\theta - \sin\theta\tan\theta - \cos\theta\sec\theta + \cos\theta\tan\theta + \sec\theta - \tan\theta \] Using \(\sin\theta \sec\theta = \tan\theta\), \(\cos\theta\sec\theta = 1\), and \(\cos\theta\tan\theta = \sin\theta\), this becomes: \[ \tan\theta - \frac{\sin^2\theta}{\cos\theta} - 1 + \sin\theta + \frac{1}{\cos\theta} - \tan\theta \]
Step 3: Group and simplify using the Pythagorean identity.
The two \(\tan\theta\) terms cancel, leaving: \[ \frac{1-\sin^2\theta}{\cos\theta} - 1 + \sin\theta = \frac{\cos^2\theta}{\cos\theta} - 1 + \sin\theta = \cos\theta - 1 + \sin\theta \]
Step 4: Compare with the required right side.
This simplifies to \(\sin\theta + \cos\theta - 1\), which is exactly the right side of the product statement in Step 1. Since the product identity holds, the original identity is proved: hence \(\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1} = \frac{1}{\sec\theta-\tan\theta}\).