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

Question

3390 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-18T00:34:07

Below

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

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

How do you prove 1+(Cosθ + Sinθ)(Cosθ-Sinθ)= 2/sec^2 θ1+(Cosθ + Sinθ)(Cosθ-Sinθ)= 2/sec^2 θ?