If \(\csc\theta + \cot\theta = p\), prove that \(\cos\theta = \dfrac{p^2-1}{p^2+1}\).
\(\cos\theta = \dfrac{p^2-1}{p^2+1}\).
Step 1: We are given that:
\[ \csc\theta + \cot\theta = p \]
Here, \(\csc\theta = \dfrac{1}{\sin\theta}\) and \(\cot\theta = \dfrac{\cos\theta}{\sin\theta}\).
Step 2: Multiply the given expression by its conjugate \((\csc\theta - \cot\theta)\):
\[(\csc\theta + \cot\theta)(\csc\theta - \cot\theta) = \csc^2\theta - \cot^2\theta.\]
Step 3: Recall the identity:
\[ \csc^2\theta - \cot^2\theta = 1. \]
So,
\[ (p)(\csc\theta - \cot\theta) = 1. \]
Therefore,
\[ \csc\theta - \cot\theta = \dfrac{1}{p}. \]
Step 4: Now we have two equations:
\[ \csc\theta + \cot\theta = p \]
\[ \csc\theta - \cot\theta = \dfrac{1}{p} \]
Step 5: Add these two equations:
\[ (\csc\theta + \cot\theta) + (\csc\theta - \cot\theta) = p + \dfrac{1}{p} \]
\[ 2\csc\theta = p + \dfrac{1}{p} \]
\[ \csc\theta = \dfrac{p + \tfrac{1}{p}}{2} \]
Step 6: Subtract the two equations:
\[ (\csc\theta + \cot\theta) - (\csc\theta - \cot\theta) = p - \dfrac{1}{p} \]
\[ 2\cot\theta = p - \dfrac{1}{p} \]
\[ \cot\theta = \dfrac{p - \tfrac{1}{p}}{2} \]
Step 7: Use the relation:
\[ \cos\theta = \dfrac{\cot\theta}{\csc\theta}. \]
Step 8: Substitute the values of \(\cot\theta\) and \(\csc\theta\):
\[ \cos\theta = \dfrac{\dfrac{p - \tfrac{1}{p}}{2}}{\dfrac{p + \tfrac{1}{p}}{2}} \]
The 2’s cancel out, so:
\[ \cos\theta = \dfrac{p - \tfrac{1}{p}}{p + \tfrac{1}{p}} \]
Step 9: Simplify the fraction by multiplying numerator and denominator by \(p\):
\[ \cos\theta = \dfrac{p^2 - 1}{p^2 + 1} \]
Final Answer:
\[ \cos\theta = \dfrac{p^2 - 1}{p^2 + 1} \]