Question

If x= a sec `theta + b tan theta and y = a tan theta + b sec theta ,"prove that" (x^2 - y^2 )=(a^2 -b^2)`

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`