If `cosectheta=(p+q)/(p-q),` then `cot(pi/4+theta/2)=`

Options

A. `sqrt((p)/(q))`
B. `sqrt((q)/(p))`
C. `sqrt(pq)`
D. pq

Explanation

To solve the problem, we start with the given equation:

\[ \csc \theta = \frac{p + q}{p - q} \]

### Step 1: Express sin θ in terms of p and q
Since \(\csc \theta = \frac{1}{\sin \theta}\), we can rewrite the equation as:

\[ \sin \theta = \frac{p - q}{p + q} \]

### Step 2: Use the half-angle identity for sin θ
We can use the half-angle identity for sine, which states:

\[ \sin \theta = \frac{2 \tan(\frac{\theta}{2})}{1 + \tan^2(\frac{\theta}{2})} \]

Let \( t = \tan(\frac{\theta}{2}) \). Then we have:

\[ \frac{2t}{1 + t^2} = \frac{p - q}{p + q} \]

### Step 3: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:

\[ 2t(p + q) = (p - q)(1 + t^2) \]

Expanding both sides:

\[ 2tp + 2tq = p - q + pt^2 - qt^2 \]

### Step 4: Rearranging the equation
Rearranging the equation, we get:

\[ pt^2 - 2tp + (2tq + q - p) = 0 \]

This is a quadratic equation in \( t \).

### Step 5: Use the quadratic formula
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = p \), \( b = -2p \), and \( c = 2q + q - p \):

\[ t = \frac{2p \pm \sqrt{(-2p)^2 - 4p(2q + q - p)}}{2p} \]

### Step 6: Simplify the expression
This simplifies to:

\[ t = \frac{2p \pm \sqrt{4p^2 - 4p(3q - p)}}{2p} \]

### Step 7: Finding cotangent
Now, we need to find \( \cot\left(\frac{\pi}{4} + \frac{\theta}{2}\right) \). Using the formula for cotangent of a sum:

\[ \cot\left(\frac{\pi}{4} + \frac{\theta}{2}\right) = \frac{1 - \tan\left(\frac{\theta}{2}\right)}{1 + \tan\left(\frac{\theta}{2}\right)} \]

Substituting \( t = \tan\left(\frac{\theta}{2}\right) \):

\[ \cot\left(\frac{\pi}{4} + \frac{\theta}{2}\right) = \frac{1 - t}{1 + t} \]

### Step 8: Relate to p and q
From our earlier work, we found that:

\[ \frac{1 + t}{1 - t} = \frac{p}{q} \]

Thus:

\[ \cot\left(\frac{\pi}{4} + \frac{\theta}{2}\right) = \sqrt{\frac{q}{p}} \]

### Final Answer
Thus, we conclude that:

\[ \cot\left(\frac{\pi}{4} + \frac{\theta}{2}\right) = \sqrt{\frac{q}{p}} \]