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Basic Trigonometric Identities

Prove that : ...

Question

Prove that :
$\frac{s i n A + c o s A}{s i n A - c o s A} + \frac{s i n A - c o s A}{s i n A + c o s A} = \frac{2}{s i n^{2} A - c o s^{2} A} = \frac{2}{2 s i n^{2} A - 1} = \frac{2}{1 - 2 c o s^{2} A}$

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Solution

L.H.S. = $\frac{s i n A + c o s A}{s i n A - c o s A} + \frac{s i n A - c o s A}{s i n A + c o s A}$

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

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

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

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

$\Rightarrow$ L.H.S. = $\frac{2}{2 s i n^{2} A - 1} = \frac{2}{2 (1 - c o s^{2} A) - 1} = \frac{2}{1 - 2 c o s^{2} A} = R . H . S .$

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Q. 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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