If `tantheta + sectheta =p`, then the value of `sec theta` is:

Options

A. `(p^(2) +1)/p^(2)`
B. `(p^(2) +1)/sqrt(p)`
C. `(p^(2) +1)/(2p)`
D. `(p+1)/(2p)`

Answer

To find the value of \( \sec \theta \) given that \( \tan \theta + \sec \theta = p \), we can follow these steps:

### Step 1: Write down the given equation
We start with the equation:
\[ \tan \theta + \sec \theta = p \]

### Step 2: Use the identity for secant
Recall the trigonometric identity:
\[ \sec^2 \theta = 1 + \tan^2 \theta \]
We can express \( \sec \theta \) in terms of \( \tan \theta \):
\[ \sec \theta = \sqrt{1 + \tan^2 \theta} \]

### Step 3: Isolate \( \sec \theta \)
From the given equation, we can isolate \( \sec \theta \):
\[ \sec \theta = p - \tan \theta \]

### Step 4: Substitute \( \tan \theta \) in terms of \( \sec \theta \)
Using the identity \( \tan^2 \theta = \sec^2 \theta - 1 \), we can express \( \tan \theta \) as:
\[ \tan \theta = \sqrt{\sec^2 \theta - 1} \]

### Step 5: Substitute back into the equation
Now, substitute \( \tan \theta \) back into the equation:
\[ \sec \theta = p - \sqrt{\sec^2 \theta - 1} \]

### Step 6: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \sec^2 \theta = (p - \sqrt{\sec^2 \theta - 1})^2 \]
Expanding the right side:
\[ \sec^2 \theta = p^2 - 2p\sqrt{\sec^2 \theta - 1} + (\sec^2 \theta - 1) \]

### Step 7: Rearrange the equation
Rearranging gives:
\[ 2p\sqrt{\sec^2 \theta - 1} = p^2 - 1 \]
Thus,
\[ \sqrt{\sec^2 \theta - 1} = \frac{p^2 - 1}{2p} \]

### Step 8: Square again to solve for \( \sec \theta \)
Squaring both sides again:
\[ \sec^2 \theta - 1 = \left(\frac{p^2 - 1}{2p}\right)^2 \]
This simplifies to:
\[ \sec^2 \theta = 1 + \frac{(p^2 - 1)^2}{4p^2} \]

### Step 9: Solve for \( \sec \theta \)
Taking the square root gives:
\[ \sec \theta = \sqrt{1 + \frac{(p^2 - 1)^2}{4p^2}} \]

### Step 10: Final expression for \( \sec \theta \)
Finally, we can express \( \sec \theta \) as:
\[ \sec \theta = \frac{p^2 + 1}{2p} \]

### Conclusion
Thus, the value of \( \sec \theta \) is:
\[ \sec \theta = \frac{p^2 + 1}{2p} \]