$\begin{array}{l}A = \sin^2\alpha + \cot^2\alpha.\sin^2\alpha\\ =\sin^2\alpha(1 + \cot^2\alpha)\\ = \sin^2\alpha\left(\dfrac{\sin^2\alpha}{\sin^2\alpha} + \dfrac{\cos^2\alpha}{\sin^2\alpha}\right)\\ =\sin^2\alpha + \cos^2\alpha = 1\\ B = \dfrac{\tan\alpha}{\cot\alpha} + \dfrac{\cot\alpha}{\tan\alpha}\\ =\tan^2\alpha + \cot^2\alpha\\ = \dfrac{\sin^2\alpha}{\cos^2\alpha} + \dfrac{\cos^2\alpha}{\sin^2\alpha}\\ = \dfrac{\sin^4\alpha + \cos^4\alpha}{\sin^2\alpha\cos^2\alpha}\\ = \dfrac{(\sin^2\alpha + \cos^2\alpha)^2 - 2\sin^2\alpha\cos^2\alpha}{\sin^2\alpha\cos^2\alpha}\\ = \dfrac{1}{\dfrac{1}{4}(2\sin\alpha\cos\alpha)^2} - 2\\ = \dfrac{4}{\sin^22\alpha} - 2\\ \left(= \dfrac{8}{1 - \cos4\alpha} - 2\right)\\ C = \sin^4\alpha + \cos^4\alpha + 2\sin^2\alpha\cos^2\alpha\\ = (\sin^2\alpha + \cos^2\alpha)^2 = 1\end{array}$
