Question

If p = sec θ - tan θ and q = cosec θ + cot θ, then what is p + q(p - 1) equal to?
1. -1
2. 0
3. 1
4. 2

Answer 1

Correct Answer - Option 1 : -1

Given:

p = sec θ - tan θ and

q = cosec θ + cot θ

Formulae Used:

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

 

Calculation:

Let the required value be x.

⇒ x = p + q(p - 1)

⇒ x = p + qp - q     ---(1)

⇒ p = sec θ - tan θ

⇒ p = \(\frac{1-sin\ \theta}{cos\ \theta}\)       ---(2)

⇒ q = cosec θ + cot θ

⇒ q = \([\frac{1+cos\ \theta}{sin\ \theta}]\)    ----(3)

Put the value of p, q in equation (1)

⇒ x = \(\frac{1-sin\ \theta}{cos\ \theta}\) + [\(\frac{1-sin\ \theta}{cos\ \theta}\)] × [\(\frac{1+cos\ \theta}{sin\ \theta}\)] - \([\frac{1+cos\ \theta}{sin\ \theta}]\)

⇒ x = \(\frac{1-sin\ \theta}{cos\ \theta}\) - \([\frac{1+cos\ \theta}{sin\ \theta}]\) + [\(\frac{1-sin\ \theta}{cos\ \theta}\)] × [\(\frac{1+cos\ \theta}{sin\ \theta}\)] 

⇒ x = \(\frac{sin\ \theta-sin^2\ \theta-cos\ \theta-cos^2\ \theta}{sin\ \theta \ cos\ \theta}\) + \(\frac{1-sin^\ \theta+cos\ \theta-sin\ \theta \ cos\ \theta}{sin\ \theta \ cos\ \theta}\)

⇒ x = \(\frac{sin\ \theta-1-cos\ \theta+1-sin^\ \theta+cos\ \theta-sin\ \theta \ cos\ \theta}{sin\ \theta \ cos\ \theta}\)

⇒ x = \(\frac{-sin\ \theta \ cos\ \theta}{sin\ \theta \ cos\ \theta}\)

∴ x = -1