Question

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

Answer 1

Consider, secθ + cosecθ = q
⇒ [(1/sinθ) + (1/cosθ)] = q
⇒ [(sinθ + cosθ)/sinθcosθ] = q
⇒ [p/sinθcosθ] = q
⇒ sinθcosθ = p/q → (1)
Consider, sinθ + cosθ = p
Squaring on both the sides we get
(sinθ + cosθ)2 = p2
⇒ sin2θ + cos2θ + 2sinθcosθ = p2
⇒ 1 + 2(p/q) = p2 [From (1)]
⇒ (q + 2p)/q = p2
⇒ (q + 2p) = p2q
⇒ 2p = p2q – q
⇒ 2p = q(p2 – 1)
Hopefully solved

Answer 2

Question: If sinθ + cosθ = p and secθ + cosecθ = q, show that q(p² - 1) = 2p.

From the LHS we get:

$\Longrightarrow \sf q(p^{2} - 1)$

[Since the RHS has "p", which has sinθ and cosθ in it, let's try to express the RHS in terms of cosθ and sinθ so we get our desired result]

Substitute p = sinθ + cosθ and q = secθ + cosecθ.

$\Longrightarrow \sf sec\theta + cosec\theta\Bigg[\Big(sin\theta + cos\theta\Big)^{2} - 1\Bigg]$

Using (a + b)² = a² + b² + 2ab we get:

$\Longrightarrow \sf sec\theta + cosec\theta\Bigg[\Big(sin\theta\Big)^2 + \Big(cos\theta\Big)^{2} + 2\Big(sin\theta\Big)\Big(cos\theta\Big) - 1\Bigg]$

Using secθ = 1/cosθ and cosecθ = 1/sinθ we get.

$\Longrightarrow \sf \dfrac{1}{cos\theta} + \dfrac{1}{sin\theta} \Bigg[sin^2\theta + cos^2\theta + 2sin\theta cos\theta - 1\Bigg]$

Using sin²θ + cos²θ = 1 we get:

$\Longrightarrow \sf \dfrac{sin\theta + cos\theta}{sin\theta cos\theta} \Bigg[1 + 2sin\theta cos\theta - 1\Bigg]$

$\Longrightarrow \sf \dfrac{sin\theta + cos\theta}{sin\theta cos\theta} \Bigg[2sin\theta cos\theta\Bigg]$

Cancelling sinθcosθ we get:

$\Longrightarrow \sf \Big[sin\theta + cos\theta\Big]\Big[2\Big]$

We know that sinθ + cosθ = p,

$\Longrightarrow \sf 2p$

LHS = RHS

Hence Proved.