27. If x= p tan + q sec 0 and y = p sec 0 + q tan e then prove that x² - y² = q^ - p^
Answers
Answer:
x = p secθ + q tanθ and y = p tanθ + q secθ
L.H.S. = x2 - y2
= (p secθ + q tanθ)2 - (p tanθ + q secθ)2
= p2 sec2θ + 2pq secθ tanθ + q2 tan2θ - (p2tan2θ + 2pq tanθ secθ + q2sec2θ)
= p2sec2θ + 2pq secθ tanθ + q2 tan2θ - p2 tan2θ - 2pq tanθ secθ - q2 sec2θ
= (p2-q2) sec2θ + (q2-p2) tan2θ
= (p2-q2) sec2θ + (q2-p2) tan2θ = (p2-q2) (sec2θ - tan2θ)
= (p2-q2) [since 1 + tan2θ = sec2θ]
= R.H.S. ∴ x2-y2 = p2-q2.
x = p secθ + q tanθ and y = p tanθ + q secθ
L.H.S. = x2 - y2
= (p secθ + q tanθ)2 - (p tanθ + q secθ)2
= p2 sec2θ + 2pq secθ tanθ + q2 tan2θ
(p2tan2θ + 2pq tanθ secθ + q2sec2θ)
= p2sec2θ + 2pq secθ tanθ + q2 tan2θ - p2 tan2θ - 2pq tanθ secθ - q2 sec2
= (p2-q2) sec2θ + (q2-p2) tan2θ
= (p2-q2) sec2θ + (q2-p2) tan2θ = (p2-q2) (sec2θ -
tan2θ)
= (p2-q2) [since 1 + tan2θ = sec2θ]
= R.H.S. ∴ x2-y2 = p2-q2.