If `tantheta=(a)/(b)`, then `bcos2theta+asin2theta` is equal to

Options

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

Explanation

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 \( \cos 2\theta \) and \( \sin 2\theta \) in terms of \( \tan \theta \)**:
- We know the formulas:
\[ \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \]
\[ \sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} \]

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

3. **Substituting \( \cos 2\theta \) and \( \sin 2\theta \) into the expression**:
- Now substitute these values into the expression \( b \cos 2\theta + a \sin 2\theta \):
\[ b \cos 2\theta + a \sin 2\theta = b \left(\frac{b^2 - a^2}{b^2 + a^2}\right) + a \left(\frac{2ab}{a^2 + b^2}\right) \]

4. **Combine the terms**:
- The expression becomes:
\[ = \frac{b(b^2 - a^2)}{b^2 + a^2} + \frac{2a^2b}{a^2 + b^2} \]
- Since both terms have the same denominator, we can combine them:
\[ = \frac{b(b^2 - a^2) + 2a^2b}{b^2 + a^2} \]
\[ = \frac{b(b^2 - a^2 + 2a^2)}{b^2 + a^2} = \frac{b(b^2 + a^2)}{b^2 + a^2} \]

5. **Simplify the expression**:
- The \( b^2 + a^2 \) terms cancel out:
\[ = b \]

### Final Result:
Thus, we conclude that:
\[ b \cos 2\theta + a \sin 2\theta = b \]