If `sec theta - tan theta =(a+1)/(a-1),` then `cos theta =`

Options

A. `(a^(2) +1)/(a ^(2)-1)`
B. `(a^(2) -1)/(a ^(2)+1)`
C. `(2a)/(a ^(2)+1)`
D. `(2a)/(a ^(2)-1)`

Answer

To solve the problem where \( \sec \theta - \tan \theta = \frac{a+1}{a-1} \) and we need to find \( \cos \theta \), we can follow these steps:

### Step 1: Use the identity for secant and tangent
We know 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 given expression
Let \( x = \sec \theta - \tan \theta \). From the problem, we have:
\[ x = \frac{a+1}{a-1} \]
Thus, we can write:
\[ \sec \theta + \tan \theta = \frac{1}{x} = \frac{a-1}{a+1} \]

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

### Step 4: Add the two equations
Adding these two equations gives:
\[ (\sec \theta - \tan \theta) + (\sec \theta + \tan \theta) = \frac{a+1}{a-1} + \frac{a-1}{a+1} \]
This simplifies to:
\[ 2 \sec \theta = \frac{a+1}{a-1} + \frac{a-1}{a+1} \]

### Step 5: Find a common denominator
The common denominator for the right side is \((a-1)(a+1)\):
\[ \frac{(a+1)^2 + (a-1)^2}{(a-1)(a+1)} \]
Calculating the numerator:
\[ (a+1)^2 + (a-1)^2 = (a^2 + 2a + 1) + (a^2 - 2a + 1) = 2a^2 + 2 \]
Thus, we have:
\[ 2 \sec \theta = \frac{2(a^2 + 1)}{(a-1)(a+1)} \]

### Step 6: Solve for sec theta
Dividing both sides by 2 gives:
\[ \sec \theta = \frac{a^2 + 1}{(a-1)(a+1)} \]

### Step 7: Find cos theta
Since \( \sec \theta = \frac{1}{\cos \theta} \), we can write:
\[ \cos \theta = \frac{(a-1)(a+1)}{a^2 + 1} \]
This simplifies to:
\[ \cos \theta = \frac{a^2 - 1}{a^2 + 1} \]

### Final Result
Thus, the value of \( \cos \theta \) is:
\[ \cos \theta = \frac{a^2 - 1}{a^2 + 1} \]