Prove the identity: (cos θ + sin θ)/(cosθ-sinθ) ≡ sec 2θ + tan 2θ

First rewrite right hand side in terms of sinθ and cosθ, because those are the terms we'll be dealing with on the left hand side: sec2θ+tan2θ = 1/cos2θ + sin2θ/cos2θ, so RHS = (1+sin2θ)/cos2θNow look at the LHS side terms. We probably want to get rid of the cosθ-sinθ on the bottom line to try and get the LHS to look like the RHS. Try multiplying by (cosθ+sinθ) on top and bottom: gives (cos²θ+sin²θ+ 2cosθsinθ)/(cos²θ-sin²θ)Now apply double angle formulas: cos²θ+sin²θ=1 sin2θ= 2cosθsinθ cos²θ-sin²θ=cos2θsubstituting in with these formulas leaves: (1+sin2θ)/cos2θwhich, as we worked out at the start, is equal to sec2θ+tan2θ!