Prove the following identities:

$(s e c A - c o s e c A) (1 + t a n A + c o t A) = t a n A s e c A - c o t A c o s e c A$

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Solution

$L H S = (s e c A - c o s e c A) (1 + t a n A + c o t A)$

$= (\frac{1}{c o s A} - \frac{1}{s i n A}) (1 + \frac{s i n A}{c o s A} + \frac{c o s A}{s i n A})$

$= (\frac{s i n A - c o s A}{s i n A c o s A}) (1 + \frac{s i n^{2} + c o s^{2} A}{s i n A c o s A})$

$= (\frac{s i n A - c o s A}{s i n A c o s A}) (1 + \frac{1}{s i n A c o s A})$

$= (\frac{s i n A - c o s A}{s i n A c o s A}) (\frac{1 + s i n A c o s A}{s i n A c o s A})$

$= \frac{(s i n A - c o s A) (s i n A c o s A + 1)}{s i n A^{2} c o s^{2} A}$

$= \frac{s i n^{2} A c o s A + s i n A - s i n A c o s^{2} A - c o s A}{s i n A^{2} c o s^{2} A}$

$= \frac{(1 - c o s^{2} A) c o s A + s i n A - s i n A (1 - s i n^{2} A) - c o s A}{s i n A^{2} c o s^{2} A}$

$= \frac{c o s A - c o s^{3} A + s i n A - s i n A + s i n^{3} A - c o s A}{s i n A^{2} c o s^{2} A}$

$= \frac{s i n^{3} A - c o s^{3} A}{s i n A^{2} c o s^{2} A}$

$= \frac{s i n^{3} A}{s i n A^{2} c o s^{2} A} - \frac{c o s^{3} A}{s i n A^{2} c o s^{2} A}$

$= \frac{s i n A}{c o s^{2} A} - \frac{c o s A}{s i n A^{2}}$

$= t a n A s e c A - c o t A c o s e c A = R H S$

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