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

Question

4435 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-03T16:51:19

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

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)#

Answer 2 · 2018-06-04T14:51:13

#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#