If `tantheta=-(1)/(sqrt(3)),sintheta=(1)/(2)andcostheta=-(sqrt(3))/(2)`, then the principal value of `theta` will be

Options

A. `(pi)/(6)`
B. `(5pi)/(6)`
C. `(7pi)/(6)`
D. `-(pi)/(6)`

Answer

To find the principal value of \( \theta \) given that \( \tan \theta = -\frac{1}{\sqrt{3}} \), \( \sin \theta = \frac{1}{2} \), and \( \cos \theta = -\frac{\sqrt{3}}{2} \), we can follow these steps:

### Step 1: Analyze the signs of trigonometric functions
We know that:
- \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)
- \( \tan \theta \) is negative when \( \sin \theta \) and \( \cos \theta \) have opposite signs.

Given:
- \( \sin \theta = \frac{1}{2} \) (positive)
- \( \cos \theta = -\frac{\sqrt{3}}{2} \) (negative)

This means that \( \theta \) must be in the second quadrant, where \( \sin \) is positive and \( \cos \) is negative.

**Hint for Step 1:** Identify the quadrant based on the signs of sine and cosine.

### Step 2: Find the reference angle
The reference angle can be found using the sine function. We know:
\[ \sin \theta = \frac{1}{2} \]
The angle whose sine is \( \frac{1}{2} \) is \( \frac{\pi}{6} \) (30 degrees).

**Hint for Step 2:** Use the sine inverse function to find the reference angle.

### Step 3: Determine the principal value in the second quadrant
Since \( \theta \) is in the second quadrant, we can find the principal value by subtracting the reference angle from \( \pi \):
\[ \theta = \pi - \frac{\pi}{6} \]
Calculating this gives:
\[ \theta = \frac{6\pi}{6} - \frac{\pi}{6} = \frac{5\pi}{6} \]

**Hint for Step 3:** Use the formula \( \theta = \pi - \text{reference angle} \) for angles in the second quadrant.

### Step 4: Verify the values
To ensure that \( \theta = \frac{5\pi}{6} \) satisfies the original conditions:
- Calculate \( \tan \theta \):
\[ \tan \left(\frac{5\pi}{6}\right) = \frac{\sin \left(\frac{5\pi}{6}\right)}{\cos \left(\frac{5\pi}{6}\right)} = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} \]
- Check \( \sin \theta \) and \( \cos \theta \):
\[ \sin \left(\frac{5\pi}{6}\right) = \frac{1}{2}, \quad \cos \left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} \]

Both conditions are satisfied.

**Final Answer:**
The principal value of \( \theta \) is:
\[ \theta = \frac{5\pi}{6} \]