What is `sin25^@ sin35^@sec 65^@sec 55^@` equal to ?

Options

A. `-1`
B. 0
C. `1/2`
D. 1

Answer

To solve the expression \( \sin 25^\circ \cdot \sin 35^\circ \cdot \sec 65^\circ \cdot \sec 55^\circ \), we can follow these steps:

### Step 1: Rewrite the secant functions
Recall that \( \sec \theta = \frac{1}{\cos \theta} \). Therefore, we can rewrite the expression as:
\[ \sin 25^\circ \cdot \sin 35^\circ \cdot \frac{1}{\cos 65^\circ} \cdot \frac{1}{\cos 55^\circ} \]

### Step 2: Use the co-function identities
Using the identity \( \cos(90^\circ - \theta) = \sin \theta \), we can rewrite \( \cos 65^\circ \) and \( \cos 55^\circ \):
\[ \cos 65^\circ = \sin(90^\circ - 65^\circ) = \sin 25^\circ \]
\[ \cos 55^\circ = \sin(90^\circ - 55^\circ) = \sin 35^\circ \]

### Step 3: Substitute the identities back into the expression
Now, substituting these identities into our expression gives:
\[ \sin 25^\circ \cdot \sin 35^\circ \cdot \frac{1}{\sin 25^\circ} \cdot \frac{1}{\sin 35^\circ} \]

### Step 4: Simplify the expression
Notice that \( \sin 25^\circ \) in the numerator and denominator cancels out, as does \( \sin 35^\circ \):
\[ = 1 \]

### Final Result
Thus, the value of \( \sin 25^\circ \cdot \sin 35^\circ \cdot \sec 65^\circ \cdot \sec 55^\circ \) is:
\[ \boxed{1} \]