Question

Eliminate theta between the equations:

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

Answer 1

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.

Answer 2

Answer:

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.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.

Step-by-step explanation: