How do you prove #1+(Cosθ + Sinθ)(Cosθ-Sinθ)= 2/sec^2 θ#?

1 Answer
Apr 18, 2018

Below

Explanation:

#1+(costheta+sintheta)(costheta-sintheta)=2/(sectheta)^2#

LHS

#1+(costheta+sintheta)(costheta-sintheta)#

=#1+((costheta)^2-sinthetacostheta+sinthetacostheta-(sintheta)^2)#

=#1+(costheta)^2-(sintheta)^2#

=#(costheta)^2+(sintheta)^2+(costheta)^2-(sintheta)^2#

=#2(costheta)^2#

RHS

#2/(sectheta)^2#

=#2/(1/(costheta)^2#

=#2(costheta)^2#

=LHS

Therefore, #1+(costheta+sintheta)(costheta-sintheta)=2/(sectheta)^2#