prove that root secA -tanA by SecA+tanA *root cosecA -cotA by cosecA+cotA =(secA - 1) CosecA -1

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Solution

$\sqrt{\frac{s e c A - \tan A}{s e c A + \tan A}} \times \sqrt{\frac{\cos e c A - c o t A}{\cos e c A + c o t A}} = \sqrt{\frac{s e c A - t a n A}{s e c A + t a n A} \times \frac{s e c A - t a n A}{s e c A - t a n A}} \times \sqrt{\frac{c o s e c A - c o t A}{c o s e c A + c o t A} \times \frac{c o s e c A - c o t A}{c o s e c A - c o t A}} = \sqrt{\frac{{(s e c A - \tan A)}^{2}}{s e c^{2} A - t a n^{2} A}} \times \times \sqrt{\frac{{(\cos e c A - c o t A)}^{2}}{c o s e c^{2} A - c o t^{2} A}} = (s e c A - t a n A) (c o s e c A - c o t A)$

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