`cos^(2)73^(@)+cos^(2)47^(@)-sin^(2)43^(@)+sin^(2)107^(@)` is equal to

Options

A. `1//2`
B. `3//4`
C. 1
D. none of these

Explanation

To solve the expression \( \cos^2 73^\circ + \cos^2 47^\circ - \sin^2 43^\circ + \sin^2 107^\circ \), we can follow these steps:

### Step 1: Rewrite the sine terms using co-function identities
We know that:
- \( \sin(90^\circ - \theta) = \cos(\theta) \)
- \( \sin(180^\circ - \theta) = \sin(\theta) \)

Using these identities, we can rewrite \( \sin^2 43^\circ \) and \( \sin^2 107^\circ \):
- \( \sin^2 43^\circ = \cos^2(90^\circ - 43^\circ) = \cos^2 47^\circ \)
- \( \sin^2 107^\circ = \sin^2(180^\circ - 73^\circ) = \sin^2 73^\circ \)

### Step 2: Substitute the rewritten terms into the expression
Now, substitute these identities back into the original expression:
\[ \cos^2 73^\circ + \cos^2 47^\circ - \cos^2 47^\circ + \sin^2 73^\circ \]

### Step 3: Simplify the expression
Notice that \( \cos^2 47^\circ \) cancels out:
\[ \cos^2 73^\circ + \sin^2 73^\circ \]

### Step 4: Apply the Pythagorean identity
Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \cos^2 73^\circ + \sin^2 73^\circ = 1 \]

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