The value of `(sin10^(@)+sin20^(@))/(cos10^(@)+cos20^(@))` equals

Options

A. `2+sqrt(3)`
B. `sqrt(2)-1`
C. `2-sqrt(3)`
D. `sqrt(2)+1`

Answer

To solve the problem \((\sin 10^\circ + \sin 20^\circ) / (\cos 10^\circ + \cos 20^\circ)\), we will use the sum-to-product identities for sine and cosine.

### Step-by-Step Solution:

1. **Apply the Sum-to-Product Identities**:
We use the identities:
\[ \sin a + \sin b = 2 \sin\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right) \]
\[ \cos a + \cos b = 2 \cos\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right) \]
Here, let \(a = 20^\circ\) and \(b = 10^\circ\).

2. **Calculate the Sine Part**:
\[ \sin 10^\circ + \sin 20^\circ = 2 \sin\left(\frac{10^\circ + 20^\circ}{2}\right) \cos\left(\frac{20^\circ - 10^\circ}{2}\right) \]
\[ = 2 \sin(15^\circ) \cos(5^\circ) \]

3. **Calculate the Cosine Part**:
\[ \cos 10^\circ + \cos 20^\circ = 2 \cos\left(\frac{10^\circ + 20^\circ}{2}\right) \cos\left(\frac{20^\circ - 10^\circ}{2}\right) \]
\[ = 2 \cos(15^\circ) \cos(5^\circ) \]

4. **Substituting Back into the Original Expression**:
Now we substitute these results back into the original expression:
\[ \frac{\sin 10^\circ + \sin 20^\circ}{\cos 10^\circ + \cos 20^\circ} = \frac{2 \sin(15^\circ) \cos(5^\circ)}{2 \cos(15^\circ) \cos(5^\circ)} \]

5. **Cancel Common Terms**:
The \(2\) and \(\cos(5^\circ)\) cancel out:
\[ = \frac{\sin(15^\circ)}{\cos(15^\circ)} = \tan(15^\circ) \]

6. **Express \(\tan(15^\circ)\)**:
We can express \(\tan(15^\circ)\) using the tangent subtraction formula:
\[ \tan(15^\circ) = \tan(45^\circ - 30^\circ) = \frac{\tan(45^\circ) - \tan(30^\circ)}{1 + \tan(45^\circ) \tan(30^\circ)} \]
Knowing \(\tan(45^\circ) = 1\) and \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\):
\[ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \]

7. **Rationalize the Denominator**:
To rationalize:
\[ \tan(15^\circ) = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - (1)^2} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \]

### Final Answer:
Thus, the value of \(\frac{\sin 10^\circ + \sin 20^\circ}{\cos 10^\circ + \cos 20^\circ}\) is:
\[ 2 - \sqrt{3} \]