Question #04652

Question

922 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-07T22:10:23

see below

Left Hand Side:

#(tan^2x-1)/(cot^2x-1) = (sin^2x/cos^2x -1)/(cos^2x/sin^2x -1)#

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

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

#=(sin^2x-cos^2x)/cos^2x *sin^2x/(cos^2x -sin^2x)#

#=(sin^2x-cos^2x)/cos^2x *sin^2x/(-(sin^2x-cos^2x)#

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

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

#=(1-cos^2x)/(-cos^2x)#

#=1/(-cos^2x) -cos^2x/(-cos^2x)#

#=1/(-cos^2x)+cancel(cos^2x)/cancel(cos^2x)#

#=-1/cos^2x +1#

#=1-1/cos^2x#