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Eliminate theta between the equations:

q tan θ + p sec θ = x, p tan θ + q sec θ = y

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q tan θ + p sec θ = x, p tan θ + q sec θ = y

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Solution

Squaring both sides of q tan θ + p sec θ = x we get,

(q tan θ + p sec θ)2 = x2 , …………….. (A)

Now, squaring both sides of p tan θ + q sec θ = y we get,

(p tan θ + q sec θ)2 = y2, …………….. (B)

Now subtract (B) from (A) we get,

x2 - y2 = (q tan θ + p sec θ)2 - (p tan θ + q sec θ) 2

⇒ x2 - y2 = (q2 tan2 θ + p2 sec2 θ + 2qp tan θ sec θ) - (p2 tan2 θ + q2 sec2 θ + 2pq tan θ sec θ)

⇒ x2 - y2 = q2 tan2 θ + p2 sec2 θ + 2qp tan θ sec θ - p2 tan2 θ - q2 sec2 θ - 2pq tan θ sec θ

⇒ x2 - y2 = q2 tan2 θ - p2 tan2 θ + p2 sec2 θ - q2 sec2 θ

⇒ x2 - y2 = tan2 θ (q2 – p2) + sec 2 θ (p2 - q2)

⇒ x2 - y2 = - tan2 θ (p2 - q2) + sec 2 θ (p2 - q2) ⇒ x2 - y2 = sec2 θ (p2 - q2) - tan2 θ (p2 - q2)

⇒ x2 - y2 = (p2 – q2) (sec2 θ - tan2 θ)

⇒ x2 - y2 = (p2 – q2)(1), [Since sec 2 θ - tan2 θ = 1]

⇒ x2 - y2 = p2 – q2

Hence the required aliment is x2 - y2 = p2 - q2.

Squaring both sides of q tan θ + p sec θ = x we get,

(q tan θ + p sec θ)2 = x2 , …………….. (A)

Now, squaring both sides of p tan θ + q sec θ = y we get,

(p tan θ + q sec θ)2 = y2, …………….. (B)

Now subtract (B) from (A) we get,

x2 - y2 = (q tan θ + p sec θ)2 - (p tan θ + q sec θ) 2

⇒ x2 - y2 = (q2 tan2 θ + p2 sec2 θ + 2qp tan θ sec θ) - (p2 tan2 θ + q2 sec2 θ + 2pq tan θ sec θ)

⇒ x2 - y2 = q2 tan2 θ + p2 sec2 θ + 2qp tan θ sec θ - p2 tan2 θ - q2 sec2 θ - 2pq tan θ sec θ

⇒ x2 - y2 = q2 tan2 θ - p2 tan2 θ + p2 sec2 θ - q2 sec2 θ

⇒ x2 - y2 = tan2 θ (q2 – p2) + sec 2 θ (p2 - q2)

⇒ x2 - y2 = - tan2 θ (p2 - q2) + sec 2 θ (p2 - q2) ⇒ x2 - y2 = sec2 θ (p2 - q2) - tan2 θ (p2 - q2)

⇒ x2 - y2 = (p2 – q2) (sec2 θ - tan2 θ)

⇒ x2 - y2 = (p2 – q2)(1), [Since sec 2 θ - tan2 θ = 1]

⇒ x2 - y2 = p2 – q2

Hence the required aliment is x2 - y2 = p2 - q2.