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

Options

A. a
B. b
C. `b//a`
D. none

Answer

To solve the problem where \( \tan \theta = \frac{a}{b} \) and we need to find \( b \cos 2\theta + a \sin 2\theta \), we can follow these steps:

### Step 1: Express \( \sin 2\theta \) and \( \cos 2\theta \) in terms of \( \tan \theta \)
Using the double angle formulas:
- \( \sin 2\theta = \frac{2\tan \theta}{1 + \tan^2 \theta} \)
- \( \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \)

### Step 2: Substitute \( \tan \theta \) with \( \frac{a}{b} \)
Substituting \( \tan \theta = \frac{a}{b} \):
- \( \sin 2\theta = \frac{2\left(\frac{a}{b}\right)}{1 + \left(\frac{a}{b}\right)^2} = \frac{\frac{2a}{b}}{1 + \frac{a^2}{b^2}} = \frac{2a}{b + \frac{a^2}{b}} = \frac{2ab}{b^2 + a^2} \)
- \( \cos 2\theta = \frac{1 - \left(\frac{a}{b}\right)^2}{1 + \left(\frac{a}{b}\right)^2} = \frac{1 - \frac{a^2}{b^2}}{1 + \frac{a^2}{b^2}} = \frac{\frac{b^2 - a^2}{b^2}}{\frac{b^2 + a^2}{b^2}} = \frac{b^2 - a^2}{b^2 + a^2} \)

### Step 3: Substitute \( \sin 2\theta \) and \( \cos 2\theta \) into the expression
Now we substitute these into the expression \( b \cos 2\theta + a \sin 2\theta \):
\[ b \cos 2\theta + a \sin 2\theta = b \left(\frac{b^2 - a^2}{b^2 + a^2}\right) + a \left(\frac{2ab}{b^2 + a^2}\right) \]

### Step 4: Simplify the expression
Combining the terms:
\[ = \frac{b(b^2 - a^2) + 2a^2b}{b^2 + a^2} \]
\[ = \frac{b^3 - ba^2 + 2a^2b}{b^2 + a^2} \]
\[ = \frac{b^3 + a^2b}{b^2 + a^2} \]

### Step 5: Factor out common terms
Factoring out \( b \):
\[ = \frac{b(b^2 + a^2)}{b^2 + a^2} \]

### Step 6: Cancel the common terms
Since \( b^2 + a^2 \) is common in the numerator and denominator, we can cancel it:
\[ = b \]

### Final Answer
Thus, the final answer is:
\[ b \cos 2\theta + a \sin 2\theta = b \]