Question

if x=q sec 0+q tan 0 and y=p tan 0+q sec 0 then prove that x²-y²=p²-q².​

Answer 1

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.

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Step-by-step explanation:

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