If `(sin x + cos x)/(sin x - cos x) = (6)/(5)` , then the value of `(tan^(2) x + 1)/(tan^(2) x-1) ` is :

Options

A. `(35)/(61)`
B. `(61)/(60)`
C. `(60)/(61)`
D. `(61)/(35)`

Answer

To solve the equation

\[ \frac{\sin x + \cos x}{\sin x - \cos x} = \frac{6}{5}, \]

we will follow these steps:

### Step 1: Cross-Multiply
Cross-multiply the equation to eliminate the fraction:

\[ 5(\sin x + \cos x) = 6(\sin x - \cos x). \]

### Step 2: Expand Both Sides
Expanding both sides gives:

\[ 5\sin x + 5\cos x = 6\sin x - 6\cos x. \]

### Step 3: Rearrange the Equation
Rearranging the equation to group similar terms:

\[ 5\sin x - 6\sin x + 5\cos x + 6\cos x = 0, \]

which simplifies to:

\[ - \sin x + 11\cos x = 0. \]

### Step 4: Solve for \(\tan x\)
From the equation \(-\sin x + 11\cos x = 0\), we can express \(\tan x\):

\[ \sin x = 11\cos x \implies \frac{\sin x}{\cos x} = 11 \implies \tan x = 11. \]

### Step 5: Calculate \(\tan^2 x + 1\) and \(\tan^2 x - 1\)
Now, we need to find the value of

\[ \frac{\tan^2 x + 1}{\tan^2 x - 1}. \]

First, calculate \(\tan^2 x\):

\[ \tan^2 x = 11^2 = 121. \]

Now, substitute this into the expression:

\[ \frac{121 + 1}{121 - 1} = \frac{122}{120}. \]

### Step 6: Simplify the Fraction
Now simplify \(\frac{122}{120}\):

\[ \frac{122}{120} = \frac{61}{60}. \]

### Final Answer
Thus, the value of

\[ \frac{\tan^2 x + 1}{\tan^2 x - 1} = \frac{61}{60}. \]