Prove that secA-tanA 2 - 1-sinA 1-sinA

1 sinA / 1+sinA =( 1 sinA ) ( 1 sinA ) #160; /( 1+sinA ) ( 1 sinA ) #160; =( 1 sinA ) #160; ^ 2 / 1 sin ^ 2 A =( 1 sinA ) #160; ^ 2 / c ...

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Trigonometric Ratios for Sum of Two Angles

Prove that ...

Question

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

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

secA - t a n A = \frac{cos A}{1 + sin A}

\sqrt{\frac{1 + s i n A}{1 - s i n A}} = s e c A + t a n A

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

Prove the following trigonometric identities:

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

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