The value of `(cot30^(@)-cot75^(@))/(tan15^(@)-tan60^(@))` is

Options

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

Answer

To solve the expression \((\cot 30^\circ - \cot 75^\circ) / (\tan 15^\circ - \tan 60^\circ)\), we can follow these steps:

### Step 1: Identify the values of \(\cot 30^\circ\) and \(\cot 75^\circ\)

We know that:
- \(\cot 30^\circ = \frac{1}{\tan 30^\circ} = \frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3}\)
- \(\cot 75^\circ = \frac{1}{\tan 75^\circ} = \frac{1}{\tan(90^\circ - 15^\circ)} = \tan 15^\circ\)

### Step 2: Calculate \(\tan 15^\circ\)

Using the tangent subtraction formula:
\[ \tan(15^\circ) = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \]
Substituting values:
\[ \tan(15^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \]

### Step 3: Substitute values into the expression

Now substituting the values into the expression:
\[ \cot 30^\circ - \cot 75^\circ = \sqrt{3} - \tan 75^\circ \]
And since \(\tan 75^\circ = \cot 15^\circ\):
\[ \tan 15^\circ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \]

### Step 4: Calculate \(\tan 60^\circ\)

We know that:
\[ \tan 60^\circ = \sqrt{3} \]

### Step 5: Substitute into the denominator

Now substituting into the denominator:
\[ \tan 15^\circ - \tan 60^\circ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} - \sqrt{3} \]

### Step 6: Simplify the expression

Now we can simplify the entire expression:
\[ \frac{\sqrt{3} - \tan 75^\circ}{\tan 15^\circ - \sqrt{3}} \]
Using the identity \(\tan(90^\circ - \theta) = \cot \theta\), we can rewrite:
\[ = \frac{\sqrt{3} - \cot 15^\circ}{\tan 15^\circ - \sqrt{3}} \]

### Step 7: Final Calculation

After simplifying, we find that:
\[ \frac{\sqrt{3} - \tan 75^\circ}{\tan 15^\circ - \tan 60^\circ} = -1 \]

Thus, the value of the expression is:
\[ \boxed{-1} \]