If tan `theta=a/b`, then the value of b cos `2theta+asin 2 theta` is

Options

A. a
B. b
C. `b/a`
D. None of these

Answer

To solve the problem, we need to find the value of \( b \cos 2\theta + a \sin 2\theta \) given that \( \tan \theta = \frac{a}{b} \).

### Step-by-Step Solution:

1. **Express \( \sin 2\theta \) and \( \cos 2\theta \) in terms of \( \tan \theta \)**:
- We know the formulas:
\[ \sin 2\theta = \frac{2\tan \theta}{1 + \tan^2 \theta} \]
\[ \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \]
- Substituting \( \tan \theta = \frac{a}{b} \):
\[ \sin 2\theta = \frac{2 \cdot \frac{a}{b}}{1 + \left(\frac{a}{b}\right)^2} = \frac{\frac{2a}{b}}{1 + \frac{a^2}{b^2}} = \frac{2a}{b} \cdot \frac{b^2}{b^2 + a^2} = \frac{2ab}{b^2 + a^2} \]
\[ \cos 2\theta = \frac{1 - \left(\frac{a}{b}\right)^2}{1 + \left(\frac{a}{b}\right)^2} = \frac{1 - \frac{a^2}{b^2}}{1 + \frac{a^2}{b^2}} = \frac{b^2 - a^2}{b^2 + a^2} \]

2. **Substituting \( \sin 2\theta \) and \( \cos 2\theta \) into the expression**:
- We need to find:
\[ b \cos 2\theta + a \sin 2\theta \]
- Substituting the values we derived:
\[ b \cos 2\theta = b \cdot \frac{b^2 - a^2}{b^2 + a^2} = \frac{b(b^2 - a^2)}{b^2 + a^2} \]
\[ a \sin 2\theta = a \cdot \frac{2ab}{b^2 + a^2} = \frac{2a^2b}{b^2 + a^2} \]

3. **Combining the two parts**:
- Now, we combine these two results:
\[ b \cos 2\theta + a \sin 2\theta = \frac{b(b^2 - a^2) + 2a^2b}{b^2 + a^2} \]
- Simplifying the numerator:
\[ = \frac{b(b^2 - a^2 + 2a^2)}{b^2 + a^2} = \frac{b(b^2 + a^2)}{b^2 + a^2} \]

4. **Final result**:
- The expression simplifies to:
\[ = b \]

### Conclusion:
Thus, the value of \( b \cos 2\theta + a \sin 2\theta \) is \( b \).