If `cosecx=2/sqrt3` and `cotx=-1/sqrt3` for x `in[0,2pi]` then `cosx+cos2x+cos3x+....+cos100x` is

Options

A. `(1)/(2)`
B. `-(1)/(2)`
C. `-(sqrt(3))/(2)`
D. `(sqrt(3))/(2)`

Answer

To solve the problem, we need to find the value of \( \cos x + \cos 2x + \cos 3x + \ldots + \cos 100x \) given that \( \csc x = \frac{2}{\sqrt{3}} \) and \( \cot x = -\frac{1}{\sqrt{3}} \).

### Step-by-Step Solution:

1. **Find \( \sin x \) and \( \cos x \)**:
- Given \( \csc x = \frac{2}{\sqrt{3}} \), we have:
\[ \sin x = \frac{1}{\csc x} = \frac{\sqrt{3}}{2} \]
- Given \( \cot x = -\frac{1}{\sqrt{3}} \), we can find \( \cos x \) using the identity \( \cot x = \frac{\cos x}{\sin x} \):
\[ \cos x = \cot x \cdot \sin x = -\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} = -\frac{1}{2} \]

2. **Determine the quadrant for \( x \)**:
- Since \( \sin x > 0 \) and \( \cot x < 0 \), \( x \) must be in the second quadrant.

3. **Find the angle \( x \)**:
- We know \( \sin x = \frac{\sqrt{3}}{2} \) corresponds to \( x = \frac{\pi}{3} \). Since \( x \) is in the second quadrant:
\[ x = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]

4. **Calculate \( \cos 2x \) and \( \cos 3x \)**:
- Using the double angle formula:
\[ \cos 2x = 2\cos^2 x - 1 = 2\left(-\frac{1}{2}\right)^2 - 1 = 2 \cdot \frac{1}{4} - 1 = \frac{1}{2} - 1 = -\frac{1}{2} \]
- Using the triple angle formula:
\[ \cos 3x = 4\cos^3 x - 3\cos x = 4\left(-\frac{1}{2}\right)^3 - 3\left(-\frac{1}{2}\right) = 4 \cdot -\frac{1}{8} + \frac{3}{2} = -\frac{1}{2} + \frac{3}{2} = 1 \]

5. **Use the formula for the sum of cosines**:
- The formula for the sum \( S = \cos x + \cos 2x + \cos 3x + \ldots + \cos 100x \) can be expressed as:
\[ S = \frac{\sin(n \cdot b)}{\sin(b)} \cdot \cos\left(a + \frac{(n-1)b}{2}\right) \]
- Here, \( a = x \), \( b = x \), and \( n = 100 \):
\[ S = \frac{\sin(100 \cdot \frac{2\pi}{3})}{\sin(\frac{2\pi}{3})} \cdot \cos\left(\frac{2\pi}{3} + \frac{99 \cdot \frac{2\pi}{3}}{2}\right) \]

6. **Calculate \( S \)**:
- Since \( \sin(100 \cdot \frac{2\pi}{3}) = \sin(66\pi + \frac{2\pi}{3}) = \sin(\frac{2\pi}{3}) \):
\[ S = \frac{\sin(\frac{2\pi}{3})}{\sin(\frac{2\pi}{3})} \cdot \cos\left(\frac{2\pi}{3} + 33\pi\right) = 1 \cdot \cos(\frac{2\pi}{3}) = -\frac{1}{2} \]

### Final Answer:
Thus, the value of \( \cos x + \cos 2x + \cos 3x + \ldots + \cos 100x \) is \( -\frac{1}{2} \).