Question

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

Hint:

When you see higher powers or fractions, simplify denominators using identities first, then convert the whole expression to \(\sin\) and \(\cos\).

Answer 1

We need to prove:
\[ \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\csc^3 \theta}{\csc^2 \theta - 1} = \sec\theta \cdot \csc\theta\,(\sec\theta + \csc\theta) \]
Step 1: Use identities
Recall:
\[ \sec^2\theta - 1 = \tan^2\theta, \qquad \csc^2\theta - 1 = \cot^2\theta \] Substitute into the LHS:
\[ \frac{\sec^3\theta}{\tan^2\theta} + \frac{\csc^3\theta}{\cot^2\theta} \]
Step 2: Rewrite in simpler ratios
\[ \frac{\sec^3\theta}{\tan^2\theta} = \sec\theta \cdot \frac{\sec^2\theta}{\tan^2\theta} \] But: \[ \frac{\sec^2\theta}{\tan^2\theta} = \frac{1/\cos^2\theta}{\sin^2\theta/\cos^2\theta} = \frac{1}{\sin^2\theta} = \csc^2\theta \] So: \[ \frac{\sec^3\theta}{\tan^2\theta} = \sec\theta \cdot \csc^2\theta \]
Similarly: \[ \frac{\csc^3\theta}{\cot^2\theta} = \csc\theta \cdot \frac{\csc^2\theta}{\cot^2\theta} \] But: \[ \frac{\csc^2\theta}{\cot^2\theta} = \frac{1/\sin^2\theta}{\cos^2\theta/\sin^2\theta} = \frac{1}{\cos^2\theta} = \sec^2\theta \] So: \[ \frac{\csc^3\theta}{\cot^2\theta} = \csc\theta \cdot \sec^2\theta \]
Step 3: Add the expressions
LHS becomes: \[ \sec\theta \csc^2\theta + \csc\theta \sec^2\theta \] Factor common terms: \[ = \sec\theta \csc\theta (\csc\theta + \sec\theta) \]
Final Result (RHS):
\[ \boxed{ \sec\theta \cdot \csc\theta\,(\sec\theta + \csc\theta) } \]
Hence proved.