Differentiate #tan^-1((sinx-cosx)/(sinx+cosx))# with respect to #x/2#?

Question

15641 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-07T07:34:15

Let #y=tan^-1((sinx-cosx)/(sinx+cosx))#

#=>tan^-1((sinx/cosx-cosx/cosx)/(sinx/cosx+cosx/cosx))#

#=>tan^-1((tanx-1)/(tanx+1))#

#=>tan^-1((tanx-tan45^@)/(1+tanxtan45^@))#

#=>tan^-1((tan(x-45^@))#

#=> x-pi/4#

Therefore, #dy/dx= 1#

Now. let #z=x/2#

Therefore, #(dz)/dx = 1/2#

The question says, "Differentiate #tan^-1((sinx-cosx)/(sinx+cosx))# with respect to #x/2#"

#=> dy/dz = dy/dx xx dx/dz#

#=> 1 xx 2/1 =2#