Question

Prove that: \(1 + \frac{\cot^2 \alpha}{1 + \csc \alpha} = \csc \alpha\)

Hint:

Whenever you see terms like \(\cot^2\) and \(\csc\), or \(\tan^2\) and \(\sec\), immediately think of the Pythagorean identities (\(1 + \cot^2 = \csc^2\)) to create common factors.

Answer 1

Step 1: Understanding the Concept:
We use trigonometric identities to simplify the left-hand side (LHS) and match it with the right-hand side (RHS).
Step 2: Key Formula or Approach:
Use the identity: \(\cot^2 \alpha = \csc^2 \alpha - 1\).
Use algebraic identity: \(a^2 - b^2 = (a - b)(a + b)\).
Step 3: Detailed Explanation:
LHS: \(1 + \frac{\cot^2 \alpha}{1 + \csc \alpha}\)
Substitute \(\cot^2 \alpha\) with \(\csc^2 \alpha - 1\):
\[ = 1 + \frac{\csc^2 \alpha - 1}{\csc \alpha + 1} \]
Factorize the numerator using the difference of squares:
\[ = 1 + \frac{(\csc \alpha - 1)(\csc \alpha + 1)}{\csc \alpha + 1} \]
Cancel the common factor \((\csc \alpha + 1)\):
\[ = 1 + (\csc \alpha - 1) \]
\[ = 1 + \csc \alpha - 1 \]
\[ = \csc \alpha \]
This matches the RHS.
Step 4: Final Answer:
LHS = RHS. Hence proved.