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: Understanding the Question:
The topic of this question is Trigonometric Identities.
We are asked to prove a standard trigonometric identity.
The Left-Hand Side (LHS) is expressed in terms of sine and cosine, while the Right-Hand Side (RHS) is expressed in terms of secant and tangent.
To transition from sine and cosine to secant and tangent, we should divide both the numerator and the denominator of the LHS by \(\cos \theta\).

Step 2: Key Formula or Approach:
- Divide the numerator and denominator by \(\cos \theta\).
- Use the Pythagorean identity:
\[ \sec^2 \theta - \tan^2 \theta = 1 \implies (\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1 \] - We will substitute this identity into the expression to factorize and simplify the terms.

Step 3: Detailed Explanation:

• Start with the Left-Hand Side (LHS) of the identity:
\[ \text{LHS} = \frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} \]

• Divide every term in the numerator and the denominator by \(\cos \theta\):
\[ \text{LHS} = \frac{\frac{\sin \theta}{\cos \theta} - \frac{\cos \theta}{\cos \theta} + \frac{1}{\cos \theta}}{\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\cos \theta} - \frac{1}{\cos \theta}} \] \[ \text{LHS} = \frac{\tan \theta - 1 + \sec \theta}{\tan \theta + 1 - \sec \theta} \] \[ \text{LHS} = \frac{(\sec \theta + \tan \theta) - 1}{\tan \theta - \sec \theta + 1} \]

• Substitute the identity \(1 = \sec^2 \theta - \tan^2 \theta\) into the numerator:
\[ \text{LHS} = \frac{(\sec \theta + \tan \theta) - (\sec^2 \theta - \tan^2 \theta)}{\tan \theta - \sec \theta + 1} \]

• Factorize the term \((\sec^2 \theta - \tan^2 \theta)\) using the difference of squares formula:
\[ \text{LHS} = \frac{(\sec \theta + \tan \theta) - (\sec \theta - \tan \theta)(\sec \theta + \tan \theta)}{\tan \theta - \sec \theta + 1} \]

• Take out the common factor \((\sec \theta + \tan \theta)\) from the numerator:
\[ \text{LHS} = \frac{(\sec \theta + \tan \theta)[1 - (\sec \theta - \tan \theta)]}{\tan \theta - \sec \theta + 1} \] \[ \text{LHS} = \frac{(\sec \theta + \tan \theta)(1 - \sec \theta + \tan \theta)}{\tan \theta - \sec \theta + 1} \]

• Notice that the term \((1 - \sec \theta + \tan \theta)\) is identical to the denominator \(( \tan \theta - \sec \theta + 1)\). Cancel these terms:
\[ \text{LHS} = \sec \theta + \tan \theta \]

• Multiply and divide by \((\sec \theta - \tan \theta)\) to match the required Right-Hand Side (RHS):
\[ \text{LHS} = \frac{(\sec \theta + \tan \theta)(\sec \theta - \tan \theta)}{\sec \theta - \tan \theta} \] \[ \text{LHS} = \frac{\sec^2 \theta - \tan^2 \theta}{\sec \theta - \tan \theta} \] \[ \text{LHS} = \frac{1}{\sec \theta - \tan \theta} \] Since the simplified LHS matches the RHS, the identity is proven.


Step 4: Final Answer:
Hence, it is proved that \(\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta}\).