If `cosx=(-sqrt3)/2 and piltxlt(3pi)/2`, then the value of `2cot^2x+3cosec^2x` is

Options

A. 16
B. 8
C. 14
D. 18

Answer

To solve the problem step by step, we need to find the value of \(2\cot^2x + 3\csc^2x\) given that \(\cos x = -\frac{\sqrt{3}}{2}\) and \(\pi < x < \frac{3\pi}{2}\).

### Step 1: Determine the angle \(x\)
Since \(\cos x = -\frac{\sqrt{3}}{2}\) and \(x\) is in the third quadrant (as \(\pi < x < \frac{3\pi}{2}\)), we can find the reference angle. The reference angle for \(\cos\) is \(30^\circ\) (or \(\frac{\pi}{6}\)). Therefore, in the third quadrant, the angle \(x\) can be calculated as:
\[ x = \pi + \frac{\pi}{6} = \frac{6\pi}{6} + \frac{\pi}{6} = \frac{7\pi}{6} \]

### Step 2: Calculate \(\cot x\)
Using the identity \(\cot x = \frac{\cos x}{\sin x}\), we first need to find \(\sin x\). Since \(\cos x = -\frac{\sqrt{3}}{2}\), we can use the Pythagorean identity:
\[ \sin^2 x + \cos^2 x = 1 \]
Substituting \(\cos x\):
\[ \sin^2 x + \left(-\frac{\sqrt{3}}{2}\right)^2 = 1 \]
\[ \sin^2 x + \frac{3}{4} = 1 \]
\[ \sin^2 x = 1 - \frac{3}{4} = \frac{1}{4} \]
Thus, \(\sin x = -\frac{1}{2}\) (negative in the third quadrant).

Now we can find \(\cot x\):
\[ \cot x = \frac{\cos x}{\sin x} = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \sqrt{3} \]

### Step 3: Calculate \(\csc x\)
Using the identity \(\csc x = \frac{1}{\sin x}\):
\[ \csc x = \frac{1}{-\frac{1}{2}} = -2 \]

### Step 4: Calculate \(2\cot^2 x + 3\csc^2 x\)
Now we can substitute the values of \(\cot x\) and \(\csc x\) into the expression:
\[ 2\cot^2 x + 3\csc^2 x = 2(\sqrt{3})^2 + 3(-2)^2 \]
Calculating each term:
\[ = 2 \cdot 3 + 3 \cdot 4 \]
\[ = 6 + 12 = 18 \]

### Final Answer
The value of \(2\cot^2 x + 3\csc^2 x\) is \(18\).

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