Question

if x = a sec 0 and y = b tan 0 then show that b2x2-a2y2=a2b2​

Answer 1

Answer:

x=a sec 0

x/a=sec 0

x^2/a^2=sec^20

y=b tan 0

y/b= tan 0

y^2/b^2=tan^2 0

so,

x^2/a^2-y^2/b^2=sec^2 0- tan^2 0

=> (b^2x^2-a^2y^2)/a^2b^2=1

=> b^2x^2-a^2y^2=a^2b^2(proved)