If `cosx=-3/(5)andpiltxlt(3pi)/(2), then ("cosec"x+"cot"x)/("sec"x-"tan"x)` is equal to

Options

A. (a)`1/(6)`
B. (b) `-1/(3)`
C. (c)`-1/(6)`
D. (d)`2/(3)`

Answer

To solve the problem, we need to evaluate the expression \(\frac{\csc x + \cot x}{\sec x - \tan x}\) given that \(\cos x = -\frac{3}{5}\) and \(x\) lies in the interval \(\left(\pi, \frac{3\pi}{2}\right)\).

### Step 1: Determine the Quadrant
Since \(x\) is in the interval \(\left(\pi, \frac{3\pi}{2}\right)\), it lies in the third quadrant. In this quadrant, sine and cosine are negative, while tangent is positive.

### Step 2: Find Sine and Tangent Values
Given \(\cos x = -\frac{3}{5}\), we can use the Pythagorean identity to find \(\sin x\):
\[ \sin^2 x + \cos^2 x = 1 \]
Substituting the value of \(\cos x\):
\[ \sin^2 x + \left(-\frac{3}{5}\right)^2 = 1 \]
\[ \sin^2 x + \frac{9}{25} = 1 \]
\[ \sin^2 x = 1 - \frac{9}{25} = \frac{16}{25} \]
Taking the square root, we find:
\[ \sin x = -\frac{4}{5} \quad (\text{negative in the third quadrant}) \]

### Step 3: Calculate Other Trigonometric Ratios
Now we can calculate \(\sec x\), \(\csc x\), \(\tan x\), and \(\cot x\):
- \(\sec x = \frac{1}{\cos x} = \frac{1}{-\frac{3}{5}} = -\frac{5}{3}\)
- \(\csc x = \frac{1}{\sin x} = \frac{1}{-\frac{4}{5}} = -\frac{5}{4}\)
- \(\tan x = \frac{\sin x}{\cos x} = \frac{-\frac{4}{5}}{-\frac{3}{5}} = \frac{4}{3}\)
- \(\cot x = \frac{1}{\tan x} = \frac{3}{4}\)

### Step 4: Substitute Values into the Expression
Now we substitute these values into the expression \(\frac{\csc x + \cot x}{\sec x - \tan x}\):
\[ \csc x + \cot x = -\frac{5}{4} + \frac{3}{4} = -\frac{5}{4} + \frac{3}{4} = -\frac{2}{4} = -\frac{1}{2} \]
\[ \sec x - \tan x = -\frac{5}{3} - \frac{4}{3} = -\frac{5 + 4}{3} = -\frac{9}{3} = -3 \]

### Step 5: Final Calculation
Now we can compute the final value:
\[ \frac{\csc x + \cot x}{\sec x - \tan x} = \frac{-\frac{1}{2}}{-3} = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} \]

### Final Answer
Thus, the value of \(\frac{\csc x + \cot x}{\sec x - \tan x}\) is:
\[ \boxed{\frac{1}{6}} \]