Find the value of `(cos^(2)20^(@)+cos^(2)70^(@))/(sin^(2)59^(@)+sin^(2)31^(@))`.

Answer

To solve the expression \((\cos^2 20^\circ + \cos^2 70^\circ) / (\sin^2 59^\circ + \sin^2 31^\circ)\), we can follow these steps:

### Step 1: Simplify the numerator
We start with the numerator \(\cos^2 20^\circ + \cos^2 70^\circ\).

Using the identity \(\cos(90^\circ - \theta) = \sin \theta\), we can express \(\cos^2 70^\circ\) as:
\[ \cos^2 70^\circ = \sin^2(90^\circ - 70^\circ) = \sin^2 20^\circ \]
Thus, the numerator becomes:
\[ \cos^2 20^\circ + \sin^2 20^\circ \]

### Step 2: Apply the Pythagorean identity
Using the Pythagorean identity \(\cos^2 \theta + \sin^2 \theta = 1\), we have:
\[ \cos^2 20^\circ + \sin^2 20^\circ = 1 \]

### Step 3: Simplify the denominator
Now, we simplify the denominator \(\sin^2 59^\circ + \sin^2 31^\circ\).

Using the identity \(\sin(90^\circ - \theta) = \cos \theta\), we can express \(\sin^2 31^\circ\) as:
\[ \sin^2 31^\circ = \cos^2(90^\circ - 31^\circ) = \cos^2 59^\circ \]
Thus, the denominator becomes:
\[ \sin^2 59^\circ + \cos^2 59^\circ \]

### Step 4: Apply the Pythagorean identity to the denominator
Using the Pythagorean identity again, we have:
\[ \sin^2 59^\circ + \cos^2 59^\circ = 1 \]

### Step 5: Combine the results
Now we can substitute back into our original expression:
\[ \frac{\cos^2 20^\circ + \cos^2 70^\circ}{\sin^2 59^\circ + \sin^2 31^\circ} = \frac{1}{1} = 1 \]

### Final Answer
Thus, the value of the expression is:
\[ \boxed{1} \]