What is the value of `(cos 10 ^(@) + sin 20^(@))/( cos 20^(@) - sin 10 ^(@))?`

Answer

To solve the expression \((\cos 10^\circ + \sin 20^\circ) / (\cos 20^\circ - \sin 10^\circ)\), we will simplify it step by step.

### Step 1: Rewrite the trigonometric functions
We can use the identity \(\cos(90^\circ - x) = \sin x\) to rewrite \(\cos 10^\circ\) and \(\cos 20^\circ\):
\[ \cos 10^\circ = \sin(90^\circ - 10^\circ) = \sin 80^\circ \]
\[ \cos 20^\circ = \sin(90^\circ - 20^\circ) = \sin 70^\circ \]
Thus, the expression becomes:
\[ \frac{\sin 80^\circ + \sin 20^\circ}{\sin 70^\circ - \sin 10^\circ} \]

### Step 2: Apply the sum-to-product identities
Using the sum-to-product identities:
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
For the numerator:
\[ \sin 80^\circ + \sin 20^\circ = 2 \sin\left(\frac{80^\circ + 20^\circ}{2}\right) \cos\left(\frac{80^\circ - 20^\circ}{2}\right) = 2 \sin(50^\circ) \cos(30^\circ) \]
For the denominator:
\[ \sin 70^\circ - \sin 10^\circ = 2 \cos\left(\frac{70^\circ + 10^\circ}{2}\right) \sin\left(\frac{70^\circ - 10^\circ}{2}\right) = 2 \cos(40^\circ) \sin(30^\circ) \]

### Step 3: Substitute back into the expression
Now substituting back into the expression:
\[ \frac{2 \sin(50^\circ) \cos(30^\circ)}{2 \cos(40^\circ) \sin(30^\circ)} \]
The \(2\) cancels out:
\[ \frac{\sin(50^\circ) \cos(30^\circ)}{\cos(40^\circ) \sin(30^\circ)} \]

### Step 4: Simplify using known values
We know that:
\(\sin(30^\circ) = \frac{1}{2}\) and \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\). Thus, we can rewrite:
\[ \frac{\sin(50^\circ) \cdot \frac{\sqrt{3}}{2}}{\cos(40^\circ) \cdot \frac{1}{2}} = \frac{\sin(50^\circ) \sqrt{3}}{\cos(40^\circ)} \]
Since \(\sin(50^\circ) = \cos(40^\circ)\), we have:
\[ \frac{\cos(40^\circ) \sqrt{3}}{\cos(40^\circ)} = \sqrt{3} \]

### Final Answer
Thus, the value of the expression is:
\[ \sqrt{3} \]