#(1-cscx)/(1-cscx)=1# so first of all it's just multiplying #cosx/(1+cscx)# by one.
Now to look at the first part:
write it as #cosx/1*1/(1+cscx)#
Then #1+cscx=1+1/sinx :. 1/(1+cscx) = (sinx)/(sinx+1)#
Recombining with that #cos(x)# that I put aside:
#(cosx*sinx)/(sinx+1)#
#sinx/(tanx+1)# is the closest I can get.
I'm not quite sure why you say the answer should be #tanx-tanxsinx#. When I check with Wolfram http://www.wolframalpha.com/input/?i=(cosx%2F(1%2Bcscx)+ the answer isn't #tanx-tanxsinx# either. Are you sure that you've entered the question correctly?
