If sinθ + cosθ = p and secθ + cosecθ = q, then prove that q (p² - 1) = 2p

Solution:

Given, sinθ + cosθ = p and secθ + cosecθ = q

We have to prove that q (p² - 1) = 2p.

Considering LHS: q (p² - 1)

Substituting the value of p and q,

(secθ + cosecθ)[(sinθ + cosθ)² - 1]

By using algebraic identity,

(a + b)² = a² + 2ab + b²

(sinθ + cosθ)² = sin²θ + 2sinθ cosθ + cos²θ

By using trigonometric identity,

cos² A + sin² A = 1

So, sin²θ + 2sinθ cosθ + cos²θ = 1 + 2sinθ cosθ

Now, [(sinθ + cosθ)² - 1] = 1 + 2sinθ cosθ - 1

= 2sinθ cosθ

Now, (secθ + cosecθ)[(sinθ + cosθ)² - 1] = (secθ + cosecθ)(2sinθ cosθ)

We know that sec A = 1/cos A and cosec A = 1/sin A

= (1/cosθ + 1/sinθ)(2sinθ cosθ)

On simplification,

= [(sinθ + cosθ)/sinθ cosθ](2sinθ cosθ)

= 2(sinθ + cosθ)

= 2p

= RHS

LHS = RHS

Therefore, q (p² - 1) = 2p.

✦ Try This: Prove that : (cosecθ - sinθ)(secθ - cosθ) = 1/(tanθ + cotθ).

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 8

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NCERT Exemplar Class 10 Maths Exercise 8.4 Problem 10

If sinθ + cosθ = p and secθ + cosecθ = q, then prove that q (p² - 1) = 2p

Summary:

If sinθ + cosθ = p and secθ + cosecθ = q, then it is proven that q (p² - 1) = 2p

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