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

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Solution

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

L.H.S

$\frac{1 - s i n A}{1 + s i n A}$

Rationalising the denominator

$\frac{(1 - s i n A) \times (1 - s i n A)}{(1 + s i n A) (1 - s i n A)}$

$\frac{(1 - s i n A)^{2}}{(1 - s i n^{2} A)}$

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

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

$(S e c A - t a n A)^{2}$

L.H S = R H.S

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\frac{1 - sin A}{1 + sin A} = (sec A - tan A)^{2}

(sec A - tan A)^{2} = \frac{1 - sin A}{1 + sin A}

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

Prove the following trigonometric identities:

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

{(\sec A - \tan A)}^{2} = \frac{1 - \sin A}{1 + \sin A}

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