$\frac{s e c A - t a n A}{s e c A + t a n A} = 1 - 2 s e c A t a n A + 2 t a n^{2} A$

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Solution

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

LHS

Multiply above and below by (secA-tanA)

= $\frac{(s e c A - t a n A)^{2}}{(s e c^{2} A - t a n^{2} A)}$

= $(s e c^{2} A - 2 s e c A t a n A + t a n^{2} A)$

= $1 + t a n^{2} A - 2 s e c A t a n A + t a n^{2} A$

= $1 - 2 s e c A t a n A + 2 t a n^{2} A$ , proved

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