How do you verify #sin^2(-x)=tan^2x/(tan^2x+1)#?

1 Answer
Oct 27, 2016

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

#(tan^2x)/(tan^2x+1)=sin^2x#

#rArr(tan^2x)/(tan^2x+1)=sin^2(-x)#.

Explanation:

Verifying the equality of the given trigonometric expressions is determined by evaluating each side separately then comparing their results.

These trigonometric identities are used
#color(red)(sin(-x)=-sinx)#

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

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

#sin^2(-x)=(sin(-x))^2=(color(red)(-sinx))^2=sin^2x#
#color(blue)(sin^2(-x)=sin^2x" " EQ1)#

#(tan^2x)/(tan^2x+1)#

#=((color(red)(sinx/cosx))^2)/((color(red)(sinx/cosx))^2+1)#

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

#=(sin^2x/cos^2x)/(color(red)(sin^2x+cos^2x)/cos^2x)#

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

#=sin^2x/cos^2x*cos^2x/1#

#=sin^2x#

#color(blue)((tan^2x)/(tan^2x+1)=sin^2x)#

#color(blue)(sin^2(-x)=sin^2x " "EQ1)#

Therefore,

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