Question #d8d7c

Question

1199 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-04T16:58:09

#(1+tanx+secx)(1+cotx-cscx)#

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

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

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

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

#=(1+2sinxcosx-1)/(cosxsinx)#

#=(2sinxcosx)/(cosxsinx)#

#=2#