Prove that cos(α−β)cosαcosβ=1+tanαtanβ\frac{\cos(\alpha - \beta)}{\cos\alpha\cos\beta} = 1 + \tan\alpha\tan\betacosαcosβcos(α−β)=1+tanαtanβ.
False; the denominator should be sinαsinβ\sin\alpha\sin\betasinαsinβ
True; but only for acute angles
True; this is a valid identity
False; the correct identity is cos(α−β)cosαcosβ=1−tanαtanβ\frac{\cos(\alpha - \beta)}{\cos\alpha\cos\beta} = 1 - \tan\alpha\tan\betacosαcosβcos(α−β)=1−tanαtanβ