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

Step 1: Understanding the Concept:
We use fundamental trigonometric identities such as \( \sec^2 \theta - 1 = \tan^2 \theta \) and \( \text{cosec}^2 \theta - 1 = \cot^2 \theta \) to simplify the expressions.
Converting complex terms into sine and cosine usually simplifies the verification of identities.
Step 2: Key Formula or Approach:
Identity 1: \( \sec^2 \theta - 1 = \tan^2 \theta \).
Identity 2: \( \text{cosec}^2 \theta - 1 = \cot^2 \theta \).
Identity 3: \( \sec \theta = \frac{1}{\cos \theta} \), \( \text{cosec} \theta = \frac{1}{\sin \theta} \), \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), \( \cot \theta = \frac{\cos \theta}{\sin \theta} \).
Step 3: Detailed Explanation:
LHS \( = \frac{\sec^3 \theta}{\tan^2 \theta} + \frac{\text{cosec}^3 \theta}{\cot^2 \theta} \).
Converting to sine and cosine:
\[ \text{LHS} = \frac{1/\cos^3 \theta}{\sin^2 \theta / \cos^2 \theta} + \frac{1/\sin^3 \theta}{\cos^2 \theta / \sin^2 \theta} \]
Simplifying each term:
\[ \text{LHS} = \frac{1}{\cos \theta \sin^2 \theta} + \frac{1}{\sin \theta \cos^2 \theta} \]
Taking the LCM of the denominators, which is \( \sin^2 \theta \cos^2 \theta \):
\[ \text{LHS} = \frac{\cos \theta + \sin \theta}{\sin^2 \theta \cos^2 \theta} \]
Now, let's look at the RHS:
\[ \text{RHS} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \]
\[ \text{RHS} = \frac{1}{\cos \theta \sin \theta} \left( \frac{1}{\cos \theta} + \frac{1}{\sin \theta} \right) \]
\[ \text{RHS} = \frac{1}{\cos \theta \sin \theta} \left( \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} \right) \]
\[ \text{RHS} = \frac{\sin \theta + \cos \theta}{\sin^2 \theta \cos^2 \theta} \]
Since LHS \( = \) RHS, the identity is proved.
Step 4: Final Answer:
LHS \( = \) RHS. Hence Proved.