Question

If cosec θ + cot θ = p, show that (p² + 1)/(p² - 1) = sec θ​

Answer 1

Given:

• cosec θ + cos θ = p

To prove:

• (p² + 1)/(p² - 1) = sec θ

Proof:

We have,

➟ p = cosec θ + cos θ

➟ p² = cosec²θ + cot²θ + 2 cosec θ cot θ

➟ p² = 1/sin²θ + 1/tan²θ + 2/(sin θ tan θ)

As tan θ = sin θ/cos θ,

➟ p² = 1/sin²θ + cos²θ/sin²θ + 2 cos θ/sin²θ

➟ p² = (cos²θ + 2 cos θ + 1)/sin²θ

➟ p² = [(cos θ)² + 2 × (cos θ) × 1 + (1)²]/sin²θ

As sin²θ + cos²θ = 1,

➟ sin²θ = 1 - cos²θ

➟ p² = (cos θ + 1)²/(1 - cos²θ)

As a² - b² = (a + b)(a - b), so,

➟ p² = [(cos θ + 1)(cos θ + 1)]/[(1 + cos θ)(1 - cos θ)]

➟ p² = (1 + cos θ)/(1 - cos θ)

Now, taking LHS,

➟ (p² + 1)/(p² - 1)

➟ [(1 + cos θ)/(1 - cos θ) + 1] ÷ [(1 + cos θ)/(1 - cos θ) - 1]

➟ (1 + cos θ + 1 - cos θ)/(1 - cos θ) ÷ (1 + cos θ - 1 + cos θ)/(1 - cos θ)

➟ 2/(1 - cos θ) ÷ (2 cos θ)/(1 - cos θ)

➟ 2/(1 - cos θ) × (1 - cos θ)/(2 cos θ)

➟ 2/(2 cos θ)

➟ 1/cos θ

➟ sec θ [1/cos θ = sec θ]

➟ RHS

Hence, proved!

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★ Formula used:

• tan θ = sin θ/cos θ
• sin²θ + cos²θ = 1
• sec θ = 1/cos θ
• cot θ = 1/tan θ