verify the identity tan^2θ - sin^2θ = tan^2θsin^2θ

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We can start with either side.  It is always best to convert all trig functions given to sine and cosine when dealing with trig identities. I will begin on the right side. I will also use x for theta.  At the end, all you have to do is replace my x with your theta.  tan^2 x = sin^2x/cos^2x sin^2 = 1 - cos^2x [(sin^2x)/(cos^2x)]*[(1 - cos^2x)/1)] = LHS Let LHS = left hand side [sin^2x - sin^2x*cos^2x]/cos^2x = LHS sin^2x(1 - cos^2x)/cos^2x = LHS (sin^2x)/(cos^2x) - (1 - cos^2x)/1 = LHS tan^2x - sin^2x = tan^2x - sin^2x