Question 714760
On the left side we can say (sin^2)/(cos^2)-sin^2
=sin^2/cos^2-(sin^2*cos^2)/cos^2
=(sin^2 *(1-cos^2))/cos^2
={{{((sin^2(theta))/(cos^2(theta)))(1-cos^2(theta))}}}


To finish this, use {{{cos^2(theta)+sin^2(theta)=1}}} in the form of {{{1-cos^2(theta)=sin^2(theta)}}}, and perform the substitution into that last step above.