If sinθ + cosθ = p and secθ + cosecθ = q, then prove that q (p2 – 1) = 2p.
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21
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
⇒ [(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
Answered by
34
Question: If sinθ + cosθ = p and secθ + cosecθ = q, show that q(p² - 1) = 2p.
From the LHS we get:
[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.
Using (a + b)² = a² + b² + 2ab we get:
Using secθ = 1/cosθ and cosecθ = 1/sinθ we get.
Using sin²θ + cos²θ = 1 we get:
Cancelling sinθcosθ we get:
We know that sinθ + cosθ = p,
LHS = RHS
Hence Proved.
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