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

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2 Answers
May 13, 2018

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)#

May 13, 2018

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