How do you verify #1-cos^2x=tan^2(-x)cos^2x#?

1 Answer
Nov 4, 2016

Verifying the equation of trigonometric identities is determined by applying some properties

#color(red)(tanx=sinx/cosx)#

#color(blue)(tan(-x)=-tanx)#

#color(purple)(cos^2x+sin^2x=1)#

Verifying the equality:
#tan^2(-x)cos^2x=(tan(-x))^2cos^2x#

#tan^2(-x)cos^2x=(color(blue)(-tan(x)))^2cos^2x#

#tan^2(-x)cos^2x=(-tan(x))^2cos^2x#

#tan^2(-x)cos^2x=tan^2xcos^2x#

#tan^2(-x)cos^2x=color(red)((sinx/cosx)^2cos^2x#

#tan^2(-x)cos^2x=sin^2x/cancel(cos^2x)cancel(cos^2x)#

#tan^2(-x)cos^2x=color(violet)(sin^2x#

We have
#color(purple)(cos^2x+sin^2x=1)#
#rArrcolor(violet)(sin^2x=1-cos^2x#

Therefore,
#tan^2(-x)cos^2x=color(violet)(1-cos^2x#