If `y=2/(sqrt(a^2-b^2)){tan^(-1)((a-b)/(a+b))tanx/2}` , then show that `(d^2y)/(dx^2)=(bsinx)/((a+bcosx)^2)`

Explanation

Given,`y=2/(sqrt(a^2-b^2)){tan^(-1)((a-b)/(a+b))tanx/2}.sqrt((a-b)/(a+b)).1/2sec^2(x/2)`


`dy/dx=2/(sqrt(a^2-b^2))1/(1+(a+b)/(a-b)tan^)2(x/2).sqrt((a-b)/(a+b)).1/2sec^2(x/2)`

`dy/dx=sqrt(1/(a^2-b^2) xx((a-b)/(a+b)) xx (a+b)^2).(1+tan^2(x/2))/(a((1+tan^2(x
/2)))+b((1-tan^2(x/2)))`

=`dy/dx=1/(a+bcosx)`

=`(d^2y)/(dx^2)=(bsinx)/(a+bcosx)^2`