Show that the derivative of #(sinx)/(1 - cosx)# is #-cscx#?

#sinx/(1-cosx)#

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

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

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

= #(1+cosx)/sinx#

= #cscx+cotx#

As such #d/(dx)(sinx/(1-cosx))#

= #d/(dx)(cscx+cotx)#

= #d/(dx)cscx+d/(dx)cotx#

= #-cscxcotx-csc^2x#

We can't, because it's false...

Let's use the product rule.

#d/(dx)[(sinx)/(1-cosx)]#

#= sinx * -1/(1-cosx)^2 * -(-sinx) + 1/(1-cosx) * cosx#

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

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

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

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

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

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

#= -cancel(1 - cosx)/(1-cosx)^cancel(2)#

#= -1/(1-cosx)#

But we know that #-1/sinx = -cscx#. Therefore, we claim that

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

Therefore:

#sinx/(1-cosx) = 1#

#sinx = 1 - cosx#

#sinx + cosx ne 1#

Clearly, we have found that this identity is false. More definitively, we can plot this equation to get:

#sinx + cosx#:

graph{sinx + cosx [-10, 10, -5, 5]}

As this equation is not a straight horizontal line, it cannot be equal to #1# for all #x# in #RR#.