Question

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If x = p sec θ + q tan θ and y = p tan θ + q sec θ, then prove that x2 – y2 = p2 – q2.

●I WANT IT IN NOTE BOOK. ​

Answer 1

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

= p²sec²θ + 2pq secθ tanθ + q² tan²θ - (p²tan²θ + 2pq tanθ secθ + q²sec²θ)

= p²sec²θ + 2pq secθ tanθ + q²tan²θ - 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

Answer 2

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

= p²sec²θ + 2pq secθ tanθ + q² tan²θ - (p²tan²θ + 2pq tanθ secθ + q²sec²θ)

= p²sec²θ + 2pq secθ tanθ + q²tan²θ - 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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