How do you prove $\frac{sec A - 1}{{sin}^{2} A} = \frac{{sec}^{2} A}{1 + sec A}$ $(secA - 1) / (sin^2A) = (sec^2A) / (1 + secA)$ ?

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How do you prove $\frac{sec A - 1}{{sin}^{2} A} = \frac{{sec}^{2} A}{1 + sec A}$ $(secA - 1) / (sin^2A) = (sec^2A) / (1 + secA)$ ?

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Answer 1 · 2018-03-15T08:09:33

To prove,we take
${sec}^{2} A - 1 = {tan}^{2} A, and tan A = \frac{sin A}{cos A}, \frac{1}{cos A} = sec A$ $sec^2A-1=tan^2A,andtanA=sinA/cosA,1/cosA=secA$

Explanation:

$\frac{sec A - 1}{{sin}^{2} A} = \frac{{sec}^{2} A}{1 + sec A}$ $(secA-1)/sin^2A=sec^2A/(1+secA)$
$L H S = \frac{sec A - 1}{{sin}^{2} A} = \frac{(sec A - 1) (sec A + 1)}{({sin}^{2} A) (sec A + 1)}$ $LHS=(secA-1)/sin^2A=((secA-1)(secA+1))/((sin^2A)(secA+1))$
$= \frac{{sec}^{2} A - 1}{({sin}^{2} A) (sec A + 1)} = \frac{{tan}^{2} A}{{sin}^{2} A (sec A + 1)} = \frac{\frac{{sin}^{2} A}{{cos}^{2} A}}{{sin}^{2} A (sec A + 1)} = \frac{\frac{1}{{cos}^{2} A}}{sec A + 1} = \frac{{sec}^{2} A}{1 + sec A} = R H S$ $=(sec^2A-1)/((sin^2A)(secA+1))=tan^2A/(sin^2A(secA+1))=(sin^2A/cos^2A)/(sin^2A(secA+1))=(1/cos^2A)/(secA+1)=sec^2A/(1+secA)=RHS$

Answer 2 · 2018-03-17T05:55:15

Kindly refer to a Proof in the Explanation.

Explanation:

We have, ${sec}^{2} A - 1 = {tan}^{2} A = {sin}^{2} A \cdot \frac{1}{{cos}^{2} A}$ $sec^2A-1=tan^2A=sin^2A*1/cos^2A$ .

$:. (secA+1)(secA-1)=sin^2A*sec^2A$ .

$rArr (secA-1)/sin^2A=sec^2A/(1+secA$ , as desired!

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How do you prove $\frac{sec A - 1}{{sin}^{2} A} = \frac{{sec}^{2} A}{1 + sec A}$ $(secA - 1) / (sin^2A) = (sec^2A) / (1 + secA)$ ?

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Explanation:

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Explanation:

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How do you prove $\frac{sec A - 1}{{sin}^{2} A} = \frac{{sec}^{2} A}{1 + sec A}$ $(secA - 1) / (sin^2A) = (sec^2A) / (1 + secA)$ ?