Can someone help verify this trig identity? (Sinx+cosx)^2/sin^2x-cos^2x=sin^2x-cos^2x/(sinx-cosx)^2

1 Answer
Swarna Islam
Mar 19, 2018

It is verified below:

Explanation:

#(sinx+cosx)^2/(sin^2x-cos^2x)=(sin^2x-cos^2x)/(sinx-cosx)^2#

#=>(cancel((sinx+cosx))(sinx+cosx))/(cancel((sinx+cosx))(sinx-cosx))=(sin^2x-cos^2x)/(sinx-cosx)^2#

#=>((sinx+cosx)(sinx-cosx))/((sinx-cosx)(sinx-cosx))=(sin^2x-cos^2x)/(sinx-cosx)^2#

#=>color(green)((sin^2x-cos^2x)/(sinx-cosx)^2)=(sin^2x-cos^2x)/(sinx-cosx)^2#

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