If y=x+tanx then prove that cos^2x(y2)-2y+2x=0 where y2 is the second derivative...plz help me out.

Answer 1

y=x+tan(x)
y'=1+sec(x)^2
y"=0+2sec(x)*(-1/cos²(x))(-sin(x))
=sec²(x)tan(x)

=>
cos²(x)y"-2y+2x
=2cos²(x)sec²(x)tan(x)-2(x+tan(x)+2x
=2tan(x)-2x-2tan(x)+2x
=0

Answer 2

Should read:

y"=0+2sec(x)*(-1/cos²(x))(-sin(x))
=2sec²(x)tan(x)