Question

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

Answer 1

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`