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 simplify each term separately using standard trigonometric identities instead of converting everything at once.
Step 2: Key Identities Used:
sec²θ − 1 = tan²θ
csc²θ − 1 = cot²θ
tan²θ = sin²θ / cos²θ
cot²θ = cos²θ / sin²θ
Step 3: Detailed Explanation:
Given LHS:
sec³θ / tan²θ + csc³θ / cot²θ

First term:
sec³θ / tan²θ
= secθ · (sec²θ / tan²θ)

Now,
sec²θ / tan²θ
= (1/cos²θ) ÷ (sin²θ/cos²θ)
= 1 / sin²θ

So first term becomes:
secθ / sin²θ
= 1 / (cosθ sin²θ)

Second term:
csc³θ / cot²θ
= cscθ · (csc²θ / cot²θ)

Now,
csc²θ / cot²θ
= (1/sin²θ) ÷ (cos²θ/sin²θ)
= 1 / cos²θ

So second term becomes:
cscθ / cos²θ
= 1 / (sinθ cos²θ)

Now add both terms:
1/(cosθ sin²θ) + 1/(sinθ cos²θ)

Take common denominator sin²θ cos²θ:
= (sinθ + cosθ) / (sin²θ cos²θ)

Now RHS:
secθ cscθ (secθ + cscθ)
= (1/cosθ)(1/sinθ) (1/cosθ + 1/sinθ)
= 1/(sinθ cosθ) · (sinθ + cosθ)/(sinθ cosθ)
= (sinθ + cosθ)/(sin²θ cos²θ)

Thus,
LHS = RHS
Final Answer:
Hence proved using an alternative simplification method.