Prove that sin(4θ)1+cos(4θ)=tan(2θ)\frac{\sin(4\theta)}{1 + \cos(4\theta)} = \tan(2\theta)1+cos(4θ)sin(4θ)=tan(2θ).
False; the identity only holds for certain values of θ\thetaθ
True; this is a valid identity for all θ\thetaθ where defined
True; but requires θ≠0\theta \neq 0θ=0
False; the correct form is sin(2θ)1+cos(2θ)\frac{\sin(2\theta)}{1 + \cos(2\theta)}1+cos(2θ)sin(2θ)