If `tan theta=a/b` then `b cos 2theta+asin 2theta=`

Options

A. a
B. b
C. `b//a`
D. `a//b`

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. **Use the given information**: We know that \( \tan \theta = \frac{a}{b} \).

2. **Express \( \cos 2\theta \) and \( \sin 2\theta \) in terms of \( \tan \theta \)**:
- The identities for \( \cos 2\theta \) and \( \sin 2\theta \) in terms of \( \tan \theta \) are:
\[ \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \]
\[ \sin 2\theta = \frac{2\tan \theta}{1 + \tan^2 \theta} \]

3. **Substituting \( \tan \theta \)**:
- Substitute \( \tan \theta = \frac{a}{b} \) into the identities:
\[ \cos 2\theta = \frac{1 - \left(\frac{a}{b}\right)^2}{1 + \left(\frac{a}{b}\right)^2} = \frac{1 - \frac{a^2}{b^2}}{1 + \frac{a^2}{b^2}} = \frac{b^2 - a^2}{b^2 + a^2} \]
\[ \sin 2\theta = \frac{2\left(\frac{a}{b}\right)}{1 + \left(\frac{a}{b}\right)^2} = \frac{\frac{2a}{b}}{1 + \frac{a^2}{b^2}} = \frac{2a}{b} \cdot \frac{b^2}{b^2 + a^2} = \frac{2ab}{b^2 + a^2} \]

4. **Substituting into the expression**:
- Now substitute \( \cos 2\theta \) and \( \sin 2\theta \) back into the expression \( b \cos 2\theta + a \sin 2\theta \):
\[ b \cos 2\theta + a \sin 2\theta = b \cdot \frac{b^2 - a^2}{b^2 + a^2} + a \cdot \frac{2ab}{b^2 + a^2} \]

5. **Simplifying the expression**:
- Combine the terms:
\[ = \frac{b(b^2 - a^2) + 2a^2b}{b^2 + a^2} \]
\[ = \frac{b^3 - ba^2 + 2a^2b}{b^2 + a^2} \]
\[ = \frac{b^3 + a^2b}{b^2 + a^2} \]

6. **Factoring out \( b \)**:
- Factor \( b \) from the numerator:
\[ = \frac{b(b^2 + a^2)}{b^2 + a^2} \]

7. **Final result**:
- The \( b^2 + a^2 \) in the numerator and denominator cancel out:
\[ = b \]

Thus, the final answer is:
\[ b \quad \text{(This is the value of } b \cos 2\theta + a \sin 2\theta\text{)} \]