Show that :1/tan3A - tanA - 1/cot3A-cotA = cot 2A

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$\frac{1}{t a n 3 A - t a n A}$ - $\frac{1}{c o t 3 A - c o t A}$ =

(\frac{1 + t a n^{2} A}{1 + c o t^{2} A}) = {(\frac{1 - t a n A}{1 - c o t A})}^{2} = t a n^{2} A

Prove that Cot A -1 / 2- sec^{2 A = cot A/ 1+tan A}

\frac{t a n A}{1 - c o t A} + \frac{c o t A}{1 - t a n A} = 1 + s e c A c o s e c A

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