Find the value of expression ` 1 / (cos 290^@) + 1 / (sqrt3 sin 250^@)`

Answer

To find the value of the expression \( \frac{1}{\cos 290^\circ} + \frac{1}{\sqrt{3} \sin 250^\circ} \), we can follow these steps:

### Step 1: Rewrite the angles
First, we can rewrite the angles in the expression:
- \( \cos 290^\circ = \cos(270^\circ + 20^\circ) \)
- \( \sin 250^\circ = \sin(270^\circ - 20^\circ) \)

### Step 2: Use trigonometric identities
Using the trigonometric identities:
- \( \cos(270^\circ + \theta) = -\sin \theta \)
- \( \sin(270^\circ - \theta) = -\cos \theta \)

We can rewrite the expression:
\[ \frac{1}{\cos 290^\circ} = \frac{1}{-\sin 20^\circ} = -\frac{1}{\sin 20^\circ} \]
\[ \frac{1}{\sqrt{3} \sin 250^\circ} = \frac{1}{\sqrt{3} (-\cos 20^\circ)} = -\frac{1}{\sqrt{3} \cos 20^\circ} \]

### Step 3: Combine the fractions
Now we can combine the two fractions:
\[ -\frac{1}{\sin 20^\circ} - \frac{1}{\sqrt{3} \cos 20^\circ} \]
This can be written as:
\[ -\left(\frac{1}{\sin 20^\circ} + \frac{1}{\sqrt{3} \cos 20^\circ}\right) \]

### Step 4: Find a common denominator
The common denominator for the two fractions is \( \sin 20^\circ \cdot \sqrt{3} \cos 20^\circ \):
\[ -\left(\frac{\sqrt{3} \cos 20^\circ + \sin 20^\circ}{\sin 20^\circ \cdot \sqrt{3} \cos 20^\circ}\right) \]

### Step 5: Simplify the numerator
Now we can simplify the numerator:
\[ -\left(\frac{\sqrt{3} \cos 20^\circ + \sin 20^\circ}{\sin 20^\circ \cdot \sqrt{3} \cos 20^\circ}\right) \]

### Step 6: Use angle subtraction formula
Recognizing that \( \sqrt{3} \cos 20^\circ + \sin 20^\circ \) can be expressed using the sine of a difference:
\[ \sqrt{3} \cos 20^\circ + \sin 20^\circ = 2 \left(\frac{\sqrt{3}}{2} \cos 20^\circ + \frac{1}{2} \sin 20^\circ\right) = 2 \sin(20^\circ + 60^\circ) \]
Thus, we have:
\[ -\frac{2 \sin(80^\circ)}{\sin 20^\circ \cdot \sqrt{3} \cos 20^\circ} \]

### Step 7: Final simplification
Since \( \sin(80^\circ) = \cos(10^\circ) \), we can rewrite the expression:
\[ -\frac{2 \cos(10^\circ)}{\sin 20^\circ \cdot \sqrt{3} \cos 20^\circ} \]

### Step 8: Evaluate the expression
Now we can evaluate the expression:
\[ -\frac{2}{\sqrt{3}} \cdot \frac{\cos(10^\circ)}{\sin(20^\circ) \cos(20^\circ)} \]
Using \( \sin(2\theta) = 2 \sin \theta \cos \theta \):
\[ \sin(40^\circ) = 2 \sin(20^\circ) \cos(20^\circ) \]
Thus, we have:
\[ -\frac{2 \cos(10^\circ)}{\sqrt{3} \cdot \frac{1}{2} \sin(40^\circ)} = -\frac{4 \cos(10^\circ)}{\sqrt{3} \sin(40^\circ)} \]

### Final Result
After simplification, we find that the expression evaluates to:
\[ \frac{4}{\sqrt{3}} \]