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

Options

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

Answer

To solve the problem, we start with the given equation:

**Given:**
\[ a \tan \theta = b \]

We need to find the value of:
\[ a \cos 2\theta + b \sin 2\theta \]

### Step 1: Express \(\tan \theta\) in terms of \(a\) and \(b\)
From the given equation, we can express \(\tan \theta\) as:
\[ \tan \theta = \frac{b}{a} \]

### Step 2: Use the double angle formulas
We will use the double angle formulas for cosine and sine:
\[ \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \]
\[ \sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} \]

### Step 3: Substitute \(\tan \theta\) into the formulas
Substituting \(\tan \theta = \frac{b}{a}\) into the formulas:

1. For \(\cos 2\theta\):
\[ \cos 2\theta = \frac{1 - \left(\frac{b}{a}\right)^2}{1 + \left(\frac{b}{a}\right)^2} = \frac{1 - \frac{b^2}{a^2}}{1 + \frac{b^2}{a^2}} = \frac{a^2 - b^2}{a^2 + b^2} \]

2. For \(\sin 2\theta\):
\[ \sin 2\theta = \frac{2 \left(\frac{b}{a}\right)}{1 + \left(\frac{b}{a}\right)^2} = \frac{\frac{2b}{a}}{1 + \frac{b^2}{a^2}} = \frac{2b}{a + \frac{b^2}{a}} = \frac{2b}{\frac{a^2 + b^2}{a}} = \frac{2ab}{a^2 + b^2} \]

### Step 4: Substitute \(\cos 2\theta\) and \(\sin 2\theta\) into the expression
Now substitute these results into the expression \(a \cos 2\theta + b \sin 2\theta\):
\[ a \cos 2\theta + b \sin 2\theta = a \left(\frac{a^2 - b^2}{a^2 + b^2}\right) + b \left(\frac{2ab}{a^2 + b^2}\right) \]

### Step 5: Simplify the expression
Combine the terms:
\[ = \frac{a(a^2 - b^2) + 2ab^2}{a^2 + b^2} \]
\[ = \frac{a^3 - ab^2 + 2ab^2}{a^2 + b^2} \]
\[ = \frac{a^3 + ab^2}{a^2 + b^2} \]

### Step 6: Factor out common terms
Factor out \(a\) from the numerator:
\[ = \frac{a(a^2 + b^2)}{a^2 + b^2} \]

### Step 7: Cancel out the common terms
Since \(a^2 + b^2\) is in both the numerator and denominator (assuming \(a^2 + b^2 \neq 0\)):
\[ = a \]

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

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