Question

Evaluate: \[ \frac{1}{\cos 290^\circ} + \frac{1}{\sqrt{3} \sin 250^\circ} \]

• A. \( \frac{\sqrt{3}}{4} \)
• B. \( \frac{4}{\sqrt{3}} \)
• C. \( \frac{2}{\sqrt{3}} \)
• D. \( \frac{\sqrt{3}}{2} \)

Hint:

When simplifying trigonometric expressions, use standard angle values or identities to relate terms to simpler trigonometric functions.

Answer 1

The Correct Option is B

We are given: \[ \frac{1}{\cos 290^\circ} + \frac{1}{\sqrt{3} \sin 250^\circ} \] Step 1: Calculate \( \cos 290^\circ \) and \( \sin 250^\circ \). - \( \cos 290^\circ = \cos (360^\circ - 70^\circ) = \cos 70^\circ \), and from standard trigonometric values, \( \cos 70^\circ = \sin 20^\circ \). - \( \sin 250^\circ = \sin (270^\circ - 20^\circ) = -\cos 20^\circ \). Step 2: Substituting these values back into the expression: \[ \frac{1}{\cos 290^\circ} + \frac{1}{\sqrt{3} \sin 250^\circ} = \frac{1}{\sin 20^\circ} + \frac{1}{-\sqrt{3} \cos 20^\circ} \] Step 3: Simplify the expression. Since \( \sin 20^\circ \approx 0.342 \) and \( \cos 20^\circ \approx 0.94 \), we can further simplify: \[ \frac{1}{\cos 290^\circ} + \frac{1}{\sqrt{3} \sin 250^\circ} \approx \frac{4}{\sqrt{3}} \] % Final Answer \[ \boxed{\frac{4}{\sqrt{3}}} \]

Answer 2

Given Expression:
\[ \frac{1}{\cos 290^\circ} + \frac{1}{\sqrt{3} \sin 250^\circ} \]

Step 1: Use angle identities
First, recall standard trigonometric values using co-function and quadrant rules:
- \( \cos 290^\circ = \cos (360^\circ - 70^\circ) = \cos (-70^\circ) = \cos 70^\circ \)
(because cosine is even: \( \cos(-\theta) = \cos \theta \))

- \( \sin 250^\circ = \sin (270^\circ - 20^\circ) = -\cos 20^\circ \)
(because \( \sin(270^\circ - \theta) = -\cos \theta \))

Step 2: Substitute and simplify
So the expression becomes:
\[ \frac{1}{\cos 70^\circ} + \frac{1}{\sqrt{3} \cdot (-\cos 20^\circ)} = \frac{1}{\cos 70^\circ} - \frac{1}{\sqrt{3} \cos 20^\circ} \]

Step 3: Use exact trigonometric values
\[ \cos 70^\circ = \sin 20^\circ, \quad \cos 20^\circ = \cos 20^\circ \]
So: \[ \frac{1}{\cos 70^\circ} = \frac{1}{\sin 20^\circ} \]

We don’t know exact values of \( \sin 20^\circ \) or \( \cos 20^\circ \) in radicals, but try to verify with a calculator:
\[ \cos 70^\circ ≈ 0.3420,\quad \cos 20^\circ ≈ 0.9397 \Rightarrow \frac{1}{0.3420} ≈ 2.924,\quad \frac{1}{\sqrt{3} \cdot 0.9397} ≈ \frac{1}{1.626} ≈ 0.615 \]
So: \[ 2.924 - 0.615 ≈ 2.309 \quad \text{BUT the correct answer is } \frac{4}{\sqrt{3}} ≈ 2.309 \]
So the expression is: \[ \boxed{\frac{4}{\sqrt{3}}} \]

Final Answer:
\[ \boxed{\frac{4}{\sqrt{3}}} \]