Question

If \(\cot \theta = \frac{b}{a}\) then \(\frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta}\)=

• A. \(\frac{b-a}{b+a}\)
• B. \(\frac{b+a}{b-a}\)
• C. \(\frac{a-b}{a+b}\)
• D. \(\frac{a+b}{a-b}\)

Answer 1

The Correct Option is B

To solve the problem, we are given that \( \cot \theta = \frac{b}{a} \), and we are asked to find the value of:

\[ \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta} \]

1. Express sin θ and cos θ in terms of a and b:
Given \( \cot \theta = \frac{b}{a} \), we can construct a right triangle where:
- Adjacent side = b
- Opposite side = a
- Hypotenuse = \( \sqrt{a^2 + b^2} \)

So, from the triangle:
\[ \sin \theta = \frac{a}{\sqrt{a^2 + b^2}}, \quad \cos \theta = \frac{b}{\sqrt{a^2 + b^2}} \]

2. Substitute into the expression:
\[ \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta} = \frac{\frac{b}{\sqrt{a^2 + b^2}} + \frac{a}{\sqrt{a^2 + b^2}}}{\frac{b}{\sqrt{a^2 + b^2}} - \frac{a}{\sqrt{a^2 + b^2}}} = \frac{b + a}{b - a} \]

3. Final simplification:
\[ \frac{b + a}{b - a} \]

Final Answer:
The correct answer is option (B): \( \frac{b + a}{b - a} \).