Question

If a sec + b tan + c = 0

p sec + q tan + r = 0

prove that

(br-qc)^2 - (pc-ar)^2 = (aq-bp)^2
....
PLZ ANSWER IT FAST

Answer 1

Step-by-step explanation:

it is given that:

asecθ + btanθ + c = 0

c = -asecθ - btanθ

c = -(a secθ + b tanθ)   --- (1)

And

p secθ + q tanθ + r = 0

r = -(p secθ + q tanθ)  --- (2)

LHS:

= (br - qc)² - (pc - ar)² = (ap - bq)²

= [b(-p secθ - q tanθ) - q(-a secθ - b tanθ)] - [p(-a secθ - b tanθ) - a( -p secθ - qtanθ)]

= -bpsecθ - bqtanθ + aqsecθ + bqtanθ - [apsecθ - bptanθ + apsecθ + aqtanθ]

= -bpsecθ - bqtanθ + aqsecθ + bqtanθ - apsecθ + bptanθ - apsecθ - aqtanθ

= -bpsecθ + aqsecθ + bptanθ - aqtanθ

= (aq - bp)²(sec²θ - tan²θ)

= (aq-bp)²(1)

= (aq-bp)²

= RHS.

Hope it helps you.

#Bebrainly

Answer 2

Solution

→ a sec∅ + b tan∅ + c = 0

→ p sec∅ + q tan∅ + r = 0

By cross multiplication method,

→ sec∅/(br - qc) = - tan∅/(ar - pc) = 1/(aq - bp)

→ sec∅ = (br - qc)/(aq - bp)

→ tan∅ = (pc - ar)/(aq - bp)

→ sec²∅ - tan²∅ = 1 [Identity]

→ [(br - qc)/(aq - bp)]² - [(pc - ar)/(aq - bp)]² = 1

→ [(br - qc)² - (pc - ar)²] = (aq - bp)²

Hence Proved