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

Question

11536 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

$L.H.S = (sinx−tanx)(cosx−cotx)$

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

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

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

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