Evaluate : `cosec10^@-sqrt3sec10^@`

Answer

To evaluate the expression \( \csc(10^\circ) - \sqrt{3} \sec(10^\circ) \), we can follow these steps:

### Step 1: Rewrite the expression in terms of sine and cosine
We know that:
\[ \csc(10^\circ) = \frac{1}{\sin(10^\circ)} \quad \text{and} \quad \sec(10^\circ) = \frac{1}{\cos(10^\circ)} \]
Thus, we can rewrite the expression as:
\[ \csc(10^\circ) - \sqrt{3} \sec(10^\circ) = \frac{1}{\sin(10^\circ)} - \sqrt{3} \cdot \frac{1}{\cos(10^\circ)} \]

### Step 2: Combine the fractions
To combine the fractions, we need a common denominator, which is \( \sin(10^\circ) \cos(10^\circ) \):
\[ \frac{1}{\sin(10^\circ)} - \frac{\sqrt{3}}{\cos(10^\circ)} = \frac{\cos(10^\circ) - \sqrt{3} \sin(10^\circ)}{\sin(10^\circ) \cos(10^\circ)} \]

### Step 3: Simplify the numerator
Now we need to simplify the numerator \( \cos(10^\circ) - \sqrt{3} \sin(10^\circ) \). We can use the sine subtraction formula:
\[ \cos(10^\circ) - \sqrt{3} \sin(10^\circ) = 2 \left( \frac{1}{2} \cos(10^\circ) - \frac{\sqrt{3}}{2} \sin(10^\circ) \right) \]
Recognizing that \( \frac{1}{2} = \cos(60^\circ) \) and \( \frac{\sqrt{3}}{2} = \sin(60^\circ) \), we can rewrite this as:
\[ 2 \left( \cos(60^\circ) \cos(10^\circ) - \sin(60^\circ) \sin(10^\circ) \right) = 2 \cos(70^\circ) \]

### Step 4: Substitute back into the expression
Now substituting back into our expression, we have:
\[ \frac{2 \cos(70^\circ)}{\sin(10^\circ) \cos(10^\circ)} \]

### Step 5: Use the identity for sine of double angles
We know that \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \). Thus:
\[ \sin(20^\circ) = 2 \sin(10^\circ) \cos(10^\circ) \]
So we can rewrite the expression as:
\[ \frac{2 \cos(70^\circ)}{\frac{1}{2} \sin(20^\circ)} = \frac{4 \cos(70^\circ)}{\sin(20^\circ)} \]

### Step 6: Evaluate \( \cos(70^\circ) \) and \( \sin(20^\circ) \)
We know that \( \cos(70^\circ) = \sin(20^\circ) \), thus:
\[ \frac{4 \sin(20^\circ)}{\sin(20^\circ)} = 4 \]

### Final Answer
Thus, the value of \( \csc(10^\circ) - \sqrt{3} \sec(10^\circ) \) is:
\[ \boxed{4} \]