Question

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

Answer 1

GIVEN:
x=psecθ+qtanθ & y=ptanθ+qsecθ

LHS = x²-y²

LHS = (psecθ+qtanθ)² - (ptanθ+qsecθ)²
LHS = (p²sec²θ+2pqsecθtanθ+q²tan²θ) - (p²tan²θ+2pqtanθsecθ+q²sec²θ)

[ (a+ b)² = a² + 2ab+ b² ]

LHS =p²sec²θ+2pqsecθtanθ+q²tan²θ - p²tan²θ -2pqsecθtanθ -q²sec²θ

LHS = p²sec²θ - q²sec²θ - p²tan²θ +q²tan²θ θ+2pq secθ tanθ -2pqsecθ tanθ
LHS = sec²θ(p²- q²)-tan²θ(p²- q²)
LHS = (p²- q²)(sec²θ - tan²θ)
LHS = (p²- q²) (1)

[ sec²θ - tan²θ =1]

x² - y² = p² - q²
LHS = RHS

HOPE THIS WILL HELP YOU....

Answer 2

Hi there!

Given :

x = p secθ + q tanθ --- (i)

y = p tanθ + q secθ ---(ii)

Squaring Eqn. (i) n' (ii)

x² = p²sec²θ + 2pq secθ.tanθ + q²tan²θ --- (iii)

y²=p²tan²θ + 2pq tanθ.secθ+ q²sec²θ ---(iv)

Subtracting Eqn. (iv) from (iii)

x² - y²

p²sec²θ + 2pq secθ.tanθ + q²tan²θ - p²tan²θ - 2pq secθ.tanθ - q²sec²θ

sec²θ (p²-q²) - tan²θ (p²-q²)

(p²-q²) × (sec²θ-tan²θ)

(p²-q²) × 1 [∵, sec²θ-tan²θ=1]

p²- q²  [ PROVED. ]

[ Thank you! for asking the question. ]
Hope it helps!