Question

Prove that d/dx ( 1 + sinx + cosx/ 1 - sinx + cosx) = secx ( secx + tanx) ??

Answer 1

Please see below.

Explanation:

We take,

`y=(1+sinx+cosx)/(1-sinx+cosx)`

#=((1+cosx)+sinx)/((1+cosx)- sinx)xx((1+cosx)+sinx)/((1+cosx)+sinx)#

`=((1+cosx+sinx)^2)/((1+cosx)^2-sin^2x)`

#=(1+cos^2x+sin^2x+2cosx+2sinx+2sinxcosx)/(1+2cosx+cos^2x- sin^2x#

`=(1+1+2cosx+2sinx+2sinxcosx)/(1-sin^2x+2cosx+cos^2x`

`=(2+2cosx+2sinx+2sinxcosx)/(cos^2x+2cosx+cos^2x)`

`=(2(1+cosx)+2sinx(1+cosx))/(2cosx+2cos^2x)`

`=(cancel((1+cosx))(2+2sinx))/(2cosxcancel((1+cosx))`

`=(2+2sinx)/(2cosx)`

`=cancel2/(cancel2cosx)+(cancel2sinx)/(cancel2cosx)`

`y=secx+tanx`

Diff.w.r.t. `x`

`(dy)/(dx)=secxtanx+sec^2x`

`=>(dy)/(dx)=secx(tanx+secx)`

Answer 2

This is one of the easiest ways to prove equalities involving derivatives. At first it may look messy but it's way easiear than the orthodox methods.Proceed step by step.

Explanation:

we have
`LHS`
`=d/dx((1+sinx+cosx)/(1-sinx+cosx))`

`=d/dx((cancel1+2sin(x/2)cos(x/2)+2cos^2(x/2)-cancel1)/(cancel1-2sin(x/2)cos(x/2)+2cos^2(x/2)-cancel1))`

`=d/dx((cancel(2cos(x/2)){cos(x/2)+sin(x/2)})/(cancel(2cos(x/2)){cos(x/2)-sin(x/2)}))`

`=d/dx((cos(x/2)+sin(x/2))/(cos(x/2)-sin(x/2)))`

`=>int(LHS)dx=intd/dx((cos(x/2)+sin(x/2))/(cos(x/2)-sin(x/2)))dx`
`=(cos(x/2)+sin(x/2))/(cos(x/2)-sin(x/2))+C ...(i)`

now
`RHS=secx(secx+tanx)`
`=>int(RHS)dx=intsecx(secx+tanx)dx`
`=>int(RHS)dx=intsec^2xdx+intsecxtanxdx`
`=tanx+secx=sinx/cosx+1/cosx=(sinx+1)/cosx`
`=(2sin(x/2)cos(x/2) + cos^2(x/2)+sin^2(x/2))/(cos^2(x/2)-sin^2(x/2))`
`=(cos(x/2)+sin(x/2))^(cancel2 1)/((cos(x/2)-sin(x/2))(cancel(cos(x/2)+sin(x/2)))`
`=(cos(x/2)+sin(x/2))/(cos(x/2)-sin(x/2))+K` ...`(ii)`
from `(i)` and `(ii)`
`intLHSdx=intRHSdx-K+C`
on differentiating both sides
`LHS=RHS`

`Q.E.D`
Here `C` and `K` are constants of integration