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how to solve (1+tan²A) / (1+cot²A) = [(1-tanA) / (1-cotA)]^{2 =} tan²A

$\frac{1 + \tan^{2} A}{1 + \cot^{2} A} = \frac{1 + \frac{\sin^{2} A}{\cos^{2} A}}{1 + \frac{\cos^{2} A}{\sin^{2} A}} = \frac{\frac{\cos^{2} A + \sin^{2} A}{\cos^{2} A}}{\frac{\sin^{2} A + \cos^{2} A}{\sin^{2} A}} = \frac{\frac{1}{\cos^{2} A}}{\frac{1}{\sin^{2} A}} = \frac{1}{\cos^{2} A} * \frac{\sin^{2} A}{1} = \tan^{2} A a n d {(\frac{1 - \tan A}{1 - \cot A})}^{2} = {(\frac{1 - \frac{\sin A}{\cos A}}{1 - \frac{\cos A}{\sin A}})}^{2} = {(\frac{\frac{\cos A - \sin A}{\cos A}}{\frac{\sin A - \cos A}{\sin A}})}^{2} = {(\frac{\frac{- (\sin A - \cos A)}{\cos A}}{\frac{\sin A - \cos A}{\sin A}})}^{2} = {(\frac{- (\sin A - \cos A)}{\sin A - \cos A} * \frac{\sin A}{\cos A})}^{2} = {(- \frac{\sin A}{\cos A})}^{2} = \frac{\sin^{2} A}{\cos^{2} A} = \tan^{2} A h e n c e p r o v e d .$

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