What is the derivative of ?

Question

#1/sinx -1/tanx#

in terms of sin and cos

519 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-30T19:43:58

#(dy)/(dx)=(1-cosx)/sin^2x#

Here,

#y=1/sinx-1/tanx#

#=>y=cscx-cotx#

We know that,

#(1)d/(dx)(cscx)=-cscxcotx#

#(2)d/(dx)(cotx)=-csc^2x#
Using #(1) and (2)# we get

#(dy)/(dx)=-cscxcotx-(-csc^2x)#

#=-cscxcotx+csc^2x#

#=csc^2x-cscxcotx#

#=1/sin^2x-1/sinx xxcosx/sinx#

#=1/sin^2x-cosx/sin^2x#

#=(1-cosx)/sin^2x#

Answer 2 · 2018-03-30T19:51:15

#(1-cosx)/sin^2x#

#d/dx(1/sinx-1/tanx)=csc^2x-cscxcotx#

#=1/sin^2x-1/sinx xxcosx/sinx#

#=(1-cosx)/sin^2x#