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 expression \(\frac{\sin 10^\circ + \sin 20^\circ}{\cos 10^\circ + \cos 20^\circ}\), we can use the sum-to-product identities for sine and cosine.

### Step-by-Step Solution:

1. **Use the Sum-to-Product Formula for Sine:**
The formula for the sum of sines is:
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
Here, let \(A = 20^\circ\) and \(B = 10^\circ\):
\[ \sin 10^\circ + \sin 20^\circ = 2 \sin\left(\frac{10^\circ + 20^\circ}{2}\right) \cos\left(\frac{20^\circ - 10^\circ}{2}\right) = 2 \sin(15^\circ) \cos(5^\circ) \]

2. **Use the Sum-to-Product Formula for Cosine:**
The formula for the sum of cosines is:
\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
Using the same \(A\) and \(B\):
\[ \cos 10^\circ + \cos 20^\circ = 2 \cos\left(\frac{10^\circ + 20^\circ}{2}\right) \cos\left(\frac{20^\circ - 10^\circ}{2}\right) = 2 \cos(15^\circ) \cos(5^\circ) \]

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

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

5. **Finding the Value of \(\tan(15^\circ)\):**
The value of \(\tan(15^\circ)\) can be calculated using the tangent subtraction formula:
\[ \tan(15^\circ) = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \]
Where \(\tan 45^\circ = 1\) and \(\tan 30^\circ = \frac{1}{\sqrt{3}}\):
\[ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \]

6. **Rationalizing the Denominator:**
To rationalize the denominator:
\[ \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - (1)^2} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \]

Thus, the final value of the expression is:
\[ \boxed{2 - \sqrt{3}} \]