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Trigonometric Identity- 1

Prove the fol...

Question

Prove the following identity.
$\frac{s i n A + c o s A}{s i n A - c o s A} + \frac{s i n A - c o s A}{s i n A + c o s A} = \frac{2}{2 s i n^{2} A - 1}$ .

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Solution

$\frac{s i n A + c o s A}{s i n A - c o s A} + \frac{s i n A - c o s A}{s i n A + c o s A} = \frac{2}{2 s i n^{2} A - 1}$
Take L.H.S
$= \frac{s i n A + c o s A}{s i n A - c o s A} + \frac{s i n A - c o s A}{s i n A + c o s A}$
$= \frac{(s i n A + c o s A)^{2} + (s i n A - c o s A)^{2}}{(s i n A - c o s A) (s i n A + c o s A)}$
$= \frac{s i n^{2} + c o s^{2} A + 2 s i n A c o s A + s i n^{2} + c o s^{2} A - 2 s i n A - c o s A}{s i n^{2} A - c o s^{2} A}$
$= \frac{1 - 1}{- c o s A} = \frac{2}{- (1 - 2 s i n^{2} A)} = \frac{2}{2 s i n^{2} A - 1}$
Hence proved

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\frac{sin A + cos A}{sin A - cos A} + \frac{sin A - cos A}{sin A + cos A} = \frac{2}{2 {sin}^{2} A - 1}

Q. Prove the identity:

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\frac{1 + cos A + sin A}{1 + cos A - sin A} = \frac{1 + sin A}{cos A}

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\frac{sin A}{1 + cos A} + \frac{1 + cos A}{sin A} = 2 c o s e c A .

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