Question

If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that (x² – y²) = (a² – b²).

Answer 1

We have `x^2 - y^2 = [( a sec theta + b tan theta )^2 - ( a tan  theta + b sec theta )^2]`

= `(a^2 sec^2 theta + b^2 tan^2 theta + 2 ab sec theta tan theta) - (a^2 tan^2 theta + b^2 sec^2 theta + 2 ab tan theta sec theta)`

= `a^2 sec^2 theta + b^2 tan^2 theta - a^2 tan^2 theta - b^2 sec^2 theta`

= `(a^2 sec^2 theta - a^2 tan^2 theta)-( b^2 sec^2 theta - b^2 tan ^2 theta)`

= `a^2 ( sec^2 theta - tan^2 theta )-b^2 ( sec^2 theta - tan^2 theta)`

= `a^2 - b^2  [∵ sec^2 theta - tan^2 theta =1]`

Hence, `x^2 - y^2 = a^2 - b^2`