How to prove (costheta+cosbeta)/(sintheta-sinbeta)=(sintheta+sinbeta)/(costheta-cosbeta ?

2 Answers
Jun 3, 2018

We use that sin^2(x)+cos^2(x)=1

Explanation:

By cross multiplication and using that
(a-b)(a+b)=a^2-b^2
we get

cos^2(theta)-cos^2(beta)=sin^2(theta)-sin^2(beta)

Jun 4, 2018

LHS=(costheta+cosbeta)/(sintheta-sinbeta)

=(costheta+cosbeta)/(sintheta-sinbeta)xx(costheta-cosbeta)/(costheta-cosbeta)

=(costheta(costheta-cosbeta)+cosbeta(costheta-cosbeta))/((sintheta-sinbeta)*(costheta-cosbeta))

=(cos^2thetacancel(-costheta*cosbeta)cancel(+costheta*cosbeta)-cos^2beta)/((sintheta-sinbeta)*(costheta-cosbeta))

=(cancel(1)-sin^2thetacancel(-1)+sin^2beta)/((sintheta-sinbeta)*(costheta-cosbeta))

=((sinbeta+sintheta)cancel((sinbeta-sintheta)))/(-cancel((sinbeta-sintheta))*(costheta-cosbeta))

=(sintheta+sinbeta)/(cosbeta-costheta)=RHS