Challenging Q

sin^-1x + cos^-1x = π/2 not 0

But i have a proof using calculus

Think about it then i will give the solution

differential of sin (arc)x is - cos (arc)x.........

Maybe we can solve........but we definitely can't prove

If u want us to prove that result, then u first prove either of the following:

0 = π/2

or

sin^-1x + cos^-1x ≠π/2

simple take y=sin^-1x
and z=cos^-1x

then dy/dx=1/√1-x²

dz/dx=-1/√1-x²

so dy/dx=-dz/dx
cancelling dx and integrating on both sides

y =-z + c

therefore

sin^-1x=-cos^-1x + c

hey one Constant should be there in ur question.......[1][1][1].....