If `sectheta+tantheta=5` , then the value of `(tantheta+1)/(tantheta-1)` is

Options

A. `(11)/(7)`
B. `(13)/(7)`
C. `(15)/(7)`
D. `(17)/(7)`

Answer

To solve the problem where \( \sec \theta + \tan \theta = 5 \) and find the value of \( \frac{\tan \theta + 1}{\tan \theta - 1} \), we can follow these steps:

### Step 1: Use the identity
We know from trigonometric identities that:
\[ \sec^2 \theta - \tan^2 \theta = 1 \]
This can be factored as:
\[ (\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1 \]

### Step 2: Substitute the known value
Given \( \sec \theta + \tan \theta = 5 \), we can substitute this into the equation:
\[ (\sec \theta - \tan \theta)(5) = 1 \]
From this, we can solve for \( \sec \theta - \tan \theta \):
\[ \sec \theta - \tan \theta = \frac{1}{5} \]

### Step 3: Set up the equations
Now we have two equations:
1. \( \sec \theta + \tan \theta = 5 \) (Equation 1)
2. \( \sec \theta - \tan \theta = \frac{1}{5} \) (Equation 2)

### Step 4: Add the two equations
Adding Equation 1 and Equation 2:
\[ (\sec \theta + \tan \theta) + (\sec \theta - \tan \theta) = 5 + \frac{1}{5} \]
This simplifies to:
\[ 2\sec \theta = 5 + \frac{1}{5} \]
Calculating the right side:
\[ 5 + \frac{1}{5} = \frac{25}{5} + \frac{1}{5} = \frac{26}{5} \]
Thus:
\[ 2\sec \theta = \frac{26}{5} \implies \sec \theta = \frac{13}{5} \]

### Step 5: Substitute to find \( \tan \theta \)
Now, substitute \( \sec \theta \) back into Equation 1 to find \( \tan \theta \):
\[ \frac{13}{5} + \tan \theta = 5 \]
Rearranging gives:
\[ \tan \theta = 5 - \frac{13}{5} = \frac{25}{5} - \frac{13}{5} = \frac{12}{5} \]

### Step 6: Calculate \( \frac{\tan \theta + 1}{\tan \theta - 1} \)
Now we can find the required expression:
\[ \frac{\tan \theta + 1}{\tan \theta - 1} = \frac{\frac{12}{5} + 1}{\frac{12}{5} - 1} \]
Convert 1 to a fraction:
\[ = \frac{\frac{12}{5} + \frac{5}{5}}{\frac{12}{5} - \frac{5}{5}} = \frac{\frac{17}{5}}{\frac{7}{5}} \]
This simplifies to:
\[ = \frac{17}{5} \cdot \frac{5}{7} = \frac{17}{7} \]

### Final Answer
Thus, the value of \( \frac{\tan \theta + 1}{\tan \theta - 1} \) is:
\[ \frac{17}{7} \]