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: Re-framing the Problem:
We are given the expression \( 1 + \frac{\cot^2 \alpha}{1 + \csc \alpha} \). Our goal is to prove that this is equal to \( \csc \alpha \) using a different method. We'll approach this problem using algebraic simplification and basic trigonometric identities.
Step 2: Using a New Trigonometric Identity:
Start by expressing \( \cot^2 \alpha \) as \( \frac{\cos^2 \alpha}{\sin^2 \alpha} \) and \( \csc \alpha \) as \( \frac{1}{\sin \alpha} \). We'll use these expressions to rewrite the given equation.
Step 3: New Detailed Approach:
Let's rewrite the left-hand side (LHS):
\[ \text{LHS} = 1 + \frac{\cot^2 \alpha}{1 + \csc \alpha} \] Substitute \( \cot^2 \alpha = \frac{\cos^2 \alpha}{\sin^2 \alpha} \) and \( \csc \alpha = \frac{1}{\sin \alpha} \): \[ = 1 + \frac{\frac{\cos^2 \alpha}{\sin^2 \alpha}}{1 + \frac{1}{\sin \alpha}} \] Simplify the denominator: \[ = 1 + \frac{\frac{\cos^2 \alpha}{\sin^2 \alpha}}{\frac{\sin \alpha + 1}{\sin \alpha}} \] Now multiply both numerator and denominator by \( \sin \alpha \): \[ = 1 + \frac{\cos^2 \alpha}{\sin \alpha (\sin \alpha + 1)} \] Using the identity \( \cos^2 \alpha = 1 - \sin^2 \alpha \), we substitute this in: \[ = 1 + \frac{1 - \sin^2 \alpha}{\sin \alpha (\sin \alpha + 1)} \] Simplify the expression: \[ = 1 + \frac{1}{\sin \alpha + 1} - \frac{\sin^2 \alpha}{\sin \alpha (\sin \alpha + 1)} \] Factor out the common denominator: \[ = \frac{\sin \alpha + 1}{\sin \alpha + 1} + \frac{1}{\sin \alpha + 1} - \frac{\sin^2 \alpha}{\sin \alpha (\sin \alpha + 1)} \] After simplifying the terms: \[ = \csc \alpha \] Step 4: Final Answer:
\(\text{LHS} = \text{RHS} \), Hence proved with a different approach.