Question

If y =cosx-sinx/cosx+sinx . Show that dy/dx +y^2+1=0

Answer 1

y =(cosx -sinx )/(cosx +sinx )

={cosx/cosx -sinx/cosx}/{cosx/cosx+sinx/cosx}

= ( 1 - tanx )/(1 +tanx)

y =(1 - tanx )/(1 +tanx)
now, differentiate wrt x

dy/dx = (1+tanx).{d(1-tanx)/dx} -(1-tanx){d(1+tanx)/dx}/(1+tanx)^2

={(1+tanx)(-sec^2x)-(1-tanx)(sec^2x) }/(1+tanx)^2

= {-2sec^2x}/(1+tanx)^2

=-2(1+tan^2)/(1+tanx)^2

=-{(1+tanx)^2+(1-tanx)^2}/(1+tanx)^2

=-(1+tanx)^2/(1+tanx)^2 -(1-tanx)^2/(1+tanx)^2

= -1 - y^2

dy/dx = -y^2 -1

dy/dx + y^2 +1 =0

//// hence proved ////

Answer 2

Step-by-step explanation:

the photograph has whole process done in it