Prove that A - tan A2 - 1 - sin A1 - sin A

LHS = ( A tan A) ^2 #160; #160; #160; #160; = (1cos A sin Acos A) ^2 #160; #160; #160; #160; =(1 sin A) ^2(1 sin ^2A) # ...

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Basic Trigonometric Identities

Prove that ...

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

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{(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:

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

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

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

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

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