Question

If x=a sec +b tan ; y=a tan +b sec prove x^2 - y^2 =a^2 -b^2

Answer 1

x=asecθ+btanθ
∴, x²=a²sec²θ+2absecθtanθ+b²tan²θ
y=atanθ+bsecθ
∴, y²=a²tan²θ+2absecθtanθ+b²sec²θ
x²-y²
=a²sec²θ+2absecθtanθ+b²tan²θ-a²tan²θ-2absecθtanθ-b²sec²θ
=a²(sec²θ-tan²θ)-b²(sec²θ-tan²θ) [∵, sec²θ-tan²θ=1]
=a²-b²

Answer 2

To Prove: x^2 - y^2 = a^2 - b^2.

Proof:

Given that,

• x = a sec A + b tan A
• y = a tan A + b sec A

By adding and subtracting x and y, we get:

(x + y) = a sec A + b tan A + a tan A + b sec A

=> (x + y) = (sec A + tan A)a + (sec A + tan A)b = (a + b)(sec A + tan A) _(1)

Similarly,

(x - y) = a sec A + b tan A - a tan A - b sec A

=> (x - y) = (sec A - tan A)a - (sec A - tan A)b = (a - b)(sec A - tan A) _(2)

Now,

Multiplying (1) w/ (2):-

(x + y)(x - y) = (a + b)(sec A + tan A)(a - b)(sec A - tan A)

=> x^2 - y^2 = a^2 - b^2(sec^2A - tan^A)

=> x^2 - y^2 = a^2 - b^2 (as sec^2A - tan^A = 1).

Hence, proved.