Question

$\csc(a ) + \cot(a) = p \: show \: that \: {p }^{2} + 1 \div {p}^{2} - 1 = \sec(a)$
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Answer 1

Step-by-step explanation:

this is proved that left hand side = right hand side

Answer 2

Required Answer:-

Given:

• cosec θ + cot θ = p

To Prove:

• (p² + 1)(p² - 1) = sec θ

Proof:

We have,

➡ p = cosec θ + cot 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 + 2cos 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)

Formula Used:

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