Question

Prove that: \(\frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\csc^3 \theta}{\csc^2 \theta - 1} = \sec \theta \cdot \csc \theta (\sec \theta + \csc \theta)\)

Hint:

When both sides look complex, simplify them independently to a basic form involving \(\sin\) and \(\cos\).

Answer 1

Step 1: Understanding the Concept:
We transform terms into \(\sin\) and \(\cos\) to simplify complex trigonometric fractions.
Step 2: Key Formula or Approach:
\(\sec^2 \theta - 1 = \tan^2 \theta\) and \(\csc^2 \theta - 1 = \cot^2 \theta\).
Step 3: Detailed Explanation:
LHS: \(\frac{\sec^3 \theta}{\tan^2 \theta} + \frac{\csc^3 \theta}{\cot^2 \theta}\)
Convert to \(\sin, \cos\):
\[ = \frac{1/\cos^3 \theta}{\sin^2 \theta/\cos^2 \theta} + \frac{1/\sin^3 \theta}{\cos^2 \theta/\sin^2 \theta} \]
\[ = \frac{1}{\cos \theta \sin^2 \theta} + \frac{1}{\sin \theta \cos^2 \theta} \]
Take common denominator \(\sin^2 \theta \cos^2 \theta\):
\[ = \frac{\sin \theta + \cos \theta}{\sin^2 \theta \cos^2 \theta} \]
RHS: \(\sec \theta \csc \theta (\sec \theta + \csc \theta)\)
\[ = \frac{1}{\cos \theta \sin \theta} \left( \frac{1}{\cos \theta} + \frac{1}{\sin \theta} \right) \]
\[ = \frac{1}{\cos \theta \sin \theta} \left( \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} \right) \]
\[ = \frac{\sin \theta + \cos \theta}{\sin^2 \theta \cos^2 \theta} \]
LHS = RHS.
Step 4: Final Answer:
Hence proved.