If `cosec theta + cot theta= p`, `cos theta = ?`

Options

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

Answer

To solve the problem where \( \csc \theta + \cot \theta = p \) and we need to find \( \cos \theta \), we can follow these steps:

### Step 1: Rewrite the given equation
We start with the equation:
\[ \csc \theta + \cot \theta = p \]
Using the definitions of cosecant and cotangent, we can rewrite this as:
\[ \frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta} = p \]
This simplifies to:
\[ \frac{1 + \cos \theta}{\sin \theta} = p \]

### Step 2: Express \(\sin \theta\) in terms of \(p\) and \(\cos \theta\)
From the equation above, we can express \(\sin \theta\):
\[ 1 + \cos \theta = p \sin \theta \]
Rearranging gives:
\[ \sin \theta = \frac{1 + \cos \theta}{p} \]

### Step 3: Use the Pythagorean identity
We know from the Pythagorean identity that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Substituting \(\sin \theta\) from Step 2 into this identity gives:
\[ \left(\frac{1 + \cos \theta}{p}\right)^2 + \cos^2 \theta = 1 \]

### Step 4: Expand and simplify
Expanding the left side:
\[ \frac{(1 + \cos \theta)^2}{p^2} + \cos^2 \theta = 1 \]
This expands to:
\[ \frac{1 + 2\cos \theta + \cos^2 \theta}{p^2} + \cos^2 \theta = 1 \]
Multiplying through by \(p^2\) to eliminate the fraction:
\[ 1 + 2\cos \theta + \cos^2 \theta + p^2 \cos^2 \theta = p^2 \]
Combining like terms:
\[ 1 + 2\cos \theta + (1 + p^2)\cos^2 \theta = p^2 \]
Rearranging gives:
\[ (1 + p^2)\cos^2 \theta + 2\cos \theta + (1 - p^2) = 0 \]

### Step 5: Solve the quadratic equation
This is a quadratic equation in \(\cos \theta\):
\[ (1 + p^2)\cos^2 \theta + 2\cos \theta + (1 - p^2) = 0 \]
Using the quadratic formula \( \cos \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \(a = 1 + p^2\), \(b = 2\), and \(c = 1 - p^2\).
\[ \cos \theta = \frac{-2 \pm \sqrt{2^2 - 4(1 + p^2)(1 - p^2)}}{2(1 + p^2)} \]
Calculating the discriminant:
\[ \sqrt{4 - 4(1 + p^2)(1 - p^2)} = \sqrt{4 - 4(1 - p^4)} = \sqrt{4p^4} \]
Thus, the equation simplifies to:
\[ \cos \theta = \frac{-2 \pm 2p^2}{2(1 + p^2)} = \frac{-1 \pm p^2}{1 + p^2} \]

### Step 6: Final result
Choosing the positive root (since cosine can be negative or positive depending on the angle):
\[ \cos \theta = \frac{p^2 - 1}{p^2 + 1} \]