Statement-1: `(cos36^(@)-cos752^(@))/(cos36^(@)cos72^(@))=2`
Statement-2: `sin15^(@)=(sqrt6-sqrt7)/(4)`

Options

A. Statement-1 is True, Statement-2 is true, Statement-2 is a correct explanation for Statement -1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To solve the problem, we need to analyze both statements and prove or disprove them step by step.

### Statement 1:
We need to prove that:
\[ \frac{\cos 36^\circ - \cos 72^\circ}{\cos 36^\circ \cos 72^\circ} = 2 \]

**Step 1: Rewrite \(\cos 72^\circ\)**
Using the identity \(\cos(90^\circ - \theta) = \sin \theta\), we can rewrite \(\cos 72^\circ\):
\[ \cos 72^\circ = \sin 18^\circ \]

**Step 2: Substitute into the equation**
Substituting this into our equation gives:
\[ \frac{\cos 36^\circ - \sin 18^\circ}{\cos 36^\circ \sin 18^\circ} \]

**Step 3: Use known values of \(\cos 36^\circ\) and \(\sin 18^\circ\)**
We know:
\[ \cos 36^\circ = \frac{\sqrt{5} + 1}{4}, \quad \sin 18^\circ = \frac{\sqrt{5} - 1}{4} \]
Substituting these values in:
\[ \frac{\frac{\sqrt{5} + 1}{4} - \frac{\sqrt{5} - 1}{4}}{\frac{\sqrt{5} + 1}{4} \cdot \frac{\sqrt{5} - 1}{4}} \]

**Step 4: Simplify the numerator**
The numerator simplifies to:
\[ \frac{\left(\sqrt{5} + 1 - (\sqrt{5} - 1)\right)}{4} = \frac{2}{4} = \frac{1}{2} \]

**Step 5: Simplify the denominator**
The denominator simplifies to:
\[ \frac{(\sqrt{5} + 1)(\sqrt{5} - 1)}{16} = \frac{5 - 1}{16} = \frac{4}{16} = \frac{1}{4} \]

**Step 6: Combine the results**
Now, substituting back into the equation:
\[ \frac{\frac{1}{2}}{\frac{1}{4}} = \frac{1}{2} \cdot 4 = 2 \]

Thus, Statement 1 is **true**.

### Statement 2:
We need to verify if:
\[ \sin 15^\circ = \frac{\sqrt{6} - \sqrt{7}}{4} \]

**Step 1: Use the sine subtraction formula**
Using the formula \(\sin(a - b) = \sin a \cos b - \cos a \sin b\):
\[ \sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ \]

**Step 2: Substitute known values**
Substituting the known values:
\[ \sin 45^\circ = \frac{1}{\sqrt{2}}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \cos 45^\circ = \frac{1}{\sqrt{2}}, \quad \sin 30^\circ = \frac{1}{2} \]
Thus, we have:
\[ \sin 15^\circ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \]

**Step 3: Simplify**
This simplifies to:
\[ \sin 15^\circ = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} - 1}{2\sqrt{2}} \]

**Step 4: Rationalize the denominator**
To rationalize:
\[ \sin 15^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{6} - \sqrt{2}}{4} \]

Since \(\frac{\sqrt{6} - \sqrt{2}}{4} \neq \frac{\sqrt{6} - \sqrt{7}}{4}\), Statement 2 is **false**.

### Conclusion:
- Statement 1 is **true**.
- Statement 2 is **false**.