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

Options

A. (a) a
B. (b) b
C. (c) `a/(b)`
D. (d) `b/(a)`

Answer

To solve the problem, we start with the given information and use trigonometric identities to find the value of the expression \( b \cos 2\theta + a \sin 2\theta \).

### Step-by-step Solution:

1. **Given**:
\[ \tan \theta = \frac{a}{b} \]

2. **Using the identity**:
\[ 1 + \tan^2 \theta = \sec^2 \theta \]
Substitute \(\tan \theta\):
\[ 1 + \left(\frac{a}{b}\right)^2 = \sec^2 \theta \]

3. **Simplifying**:
\[ 1 + \frac{a^2}{b^2} = \sec^2 \theta \]
\[ \frac{b^2 + a^2}{b^2} = \sec^2 \theta \]

4. **Finding \(\cos^2 \theta\)**:
Since \(\sec^2 \theta = \frac{1}{\cos^2 \theta}\), we have:
\[ \cos^2 \theta = \frac{b^2}{b^2 + a^2} \]

5. **Finding \(\sin^2 \theta\)**:
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\[ \sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{b^2}{b^2 + a^2} \]
\[ = \frac{(b^2 + a^2) - b^2}{b^2 + a^2} = \frac{a^2}{b^2 + a^2} \]

6. **Finding \(\sin \theta\) and \(\cos \theta\)**:
\[ \sin \theta = \sqrt{\frac{a^2}{b^2 + a^2}} = \frac{a}{\sqrt{b^2 + a^2}} \]
\[ \cos \theta = \sqrt{\frac{b^2}{b^2 + a^2}} = \frac{b}{\sqrt{b^2 + a^2}} \]

7. **Finding \(\cos 2\theta\) and \(\sin 2\theta\)**:
Using the double angle formulas:
\[ \cos 2\theta = \cos^2 \theta - \sin^2 \theta = \frac{b^2}{b^2 + a^2} - \frac{a^2}{b^2 + a^2} = \frac{b^2 - a^2}{b^2 + a^2} \]
\[ \sin 2\theta = 2 \sin \theta \cos \theta = 2 \cdot \frac{a}{\sqrt{b^2 + a^2}} \cdot \frac{b}{\sqrt{b^2 + a^2}} = \frac{2ab}{b^2 + a^2} \]

8. **Substituting into the expression**:
\[ b \cos 2\theta + a \sin 2\theta = b \cdot \frac{b^2 - a^2}{b^2 + a^2} + a \cdot \frac{2ab}{b^2 + a^2} \]
\[ = \frac{b(b^2 - a^2) + 2a^2b}{b^2 + a^2} \]
\[ = \frac{b^3 - ab^2 + 2a^2b}{b^2 + a^2} \]
\[ = \frac{b^3 + a^2b}{b^2 + a^2} \]
\[ = \frac{b(b^2 + a^2)}{b^2 + a^2} = b \]

9. **Final Result**:
Thus, the value of \( b \cos 2\theta + a \sin 2\theta \) is:
\[ \boxed{b} \]