How do you prove #(sinx - tanx)(cosx - cotx) = (sinx -1)(cosx -1)#?

Question

11017 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-03T08:13:34

#L.H.S = (sinx−tanx)(cosx−cotx)#

#=> (sinx−sinx/cosx)(cosx−cosx/sinx)#

#=> ((sinxcosx−sinx)/cosx)((cosxsinx−cosx)/sinx)#

#=> 1/(cosxsinx)(sinxcosx−sinx)(cosxsinx−cosx)#

#=> 1/(cosxsinx)(sin^2xcos^2x-cos^2xsinx-sin^2xcosx+sinxcosx)#

#=> (sinxcosx-cosx-sinx+1)#

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

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

Answer 2 · 2018-04-03T10:54:51

#L.H.S = (sinx−tanx)(cosx−cotx)#

#=tanx (sinx/tanx−tanx/tanx)*cotx(cosx/cotx−cotx/cotx)#

#=tanx*cotx (sinx/(sinx/cosx)−1)(cosx/(cosx/sinx)−1)#

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