Question

if x=a sec θand y=b tan θ then prove that b2x2-a2y2=a2b2

Answer 1

Solution:

It is given that,

x=a sec θ

=> x/a= sec θ

=> (x/a)² = sec² θ

=> x²/a² = sec²θ----(1)

and

y=b tan θ

=> y/b = tan θ

=> (y/b)² = tan²θ

=> y²/b² = tan²θ ----(2)

Subtract (2) from (1), we get

=> x²/a² - y²/b²=sec²θ-tan²θ

=> (b²x²-a²y²)/a²b² = 1

/* sec²A - tan²A = 1 */

=> b²x² - a²y² = a²b²

•••••

Answer 2

Step-by-step explanation:

Given

        x = aSecθ

      bx = abSecθ

Square on both sides

     b²x² = a²b²Sec²θ

      y = bTanθ

    ay = abTanθ

Square on both sides

   a²y² = a²b²Tan²θ

         

b²x² - a²y² = a²b²(Sec²θ - Tan²θ)

∴ b²x² - a²y² = a²b²