The most general value of `theta` which satisfies both the equation in
If `tan theta = sqrt3 and cosec theta = - 2 /sqrt3`, the most general value of `theta` satisfying both equation is

Options

A. `2n pi + pi/3`
B. `2n pi + (2pi)/3`
C. `2n pi+ (4pi)/3`
D. none

Answer

To solve the problem, we need to find the most general value of \( \theta \) that satisfies both equations:

1. \( \tan \theta = \sqrt{3} \)
2. \( \csc \theta = -\frac{2}{\sqrt{3}} \)

### Step 1: Solve \( \tan \theta = \sqrt{3} \)

The tangent function is positive in the first and third quadrants. The reference angle for \( \tan \theta = \sqrt{3} \) is \( \frac{\pi}{3} \).

Thus, the general solutions for \( \theta \) are given by:
\[ \theta = n\pi + \frac{\pi}{3} \quad \text{for } n \in \mathbb{Z} \]

This can be expressed as:
\[ \theta = n\pi + \frac{\pi}{3} \quad \text{or} \quad \theta = n\pi + \frac{4\pi}{3} \]

### Step 2: Solve \( \csc \theta = -\frac{2}{\sqrt{3}} \)

Since \( \csc \theta = \frac{1}{\sin \theta} \), we can rewrite the equation as:
\[ \sin \theta = -\frac{\sqrt{3}}{2} \]

The sine function is negative in the third and fourth quadrants. The reference angle for \( \sin \theta = -\frac{\sqrt{3}}{2} \) is \( \frac{\pi}{3} \).

Thus, the general solutions for \( \theta \) are:
\[ \theta = n\pi + \frac{4\pi}{3} \quad \text{for } n \in \mathbb{Z} \]
\[ \theta = n\pi + \frac{5\pi}{3} \quad \text{for } n \in \mathbb{Z} \]

### Step 3: Find common solutions

From the solutions derived from both equations, we have:
1. From \( \tan \theta = \sqrt{3} \):
- \( \theta = n\pi + \frac{\pi}{3} \)
- \( \theta = n\pi + \frac{4\pi}{3} \)

2. From \( \csc \theta = -\frac{2}{\sqrt{3}} \):
- \( \theta = n\pi + \frac{4\pi}{3} \)
- \( \theta = n\pi + \frac{5\pi}{3} \)

The common solution from both equations is:
\[ \theta = n\pi + \frac{4\pi}{3} \]

### Step 4: Write the general solution

The most general value of \( \theta \) satisfying both equations is:
\[ \theta = 2n\pi + \frac{4\pi}{3} \quad \text{for } n \in \mathbb{Z} \]