How do you simplify $\frac{{sec}^{2} x - 1}{{sec}^{2} x}$ $(sec^2x-1)/ sec^2x$ ?

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How do you simplify $\frac{{sec}^{2} x - 1}{{sec}^{2} x}$ $(sec^2x-1)/ sec^2x$ ?

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Answer 1 · 2015-11-02T16:15:56

Simplify $\frac{{sec}^{2} - 1}{{sec}^{2}}$ $(sec^2 - 1)/sec^2$

Ans: sin^2 x

Explanation:

Replace ${sec}^{2} = \frac{1}{{cos}^{2} x}$ $sec^2 = 1/(cos^2 x$ into the expression , we get:
$\frac{\frac{1}{{cos}^{2} x - 1}}{\frac{1}{{cos}^{2} x}} = (\frac{1 - {cos}^{2} x}{{cos}^{2} x}) (\frac{{cos}^{2} x}{1}) =$ $(1/(cos^2 x - 1))/(1/(cos^2 x)) = ((1 - cos^2 x)/cos^2 x)((cos^2 x)/1) =$

Since $(1 - {cos}^{2} x) = {sin}^{2} x$ $(1 - cos^2 x) = sin^2x$ , therefore:

$\frac{{sec}^{2} x - 1}{{sec}^{2} x} = {sin}^{2} x$ $(sec^2 x - 1)/sec^2 x = sin^2 x$

Answer 2 · 2015-11-02T16:18:20

${sin}^{2} x$ $sin^2x$

Explanation:

$1 + {tan}^{2} x = {sec}^{2} x$ $1 + tan^2x=sec^2x$

so ${sec}^{2} x - 1 = {tan}^{2} x$ $" "sec^2x-1 = tan^2x$

Giving $\frac{{tan}^{2} x}{{sec}^{2} x}$ $" " (tan^2x)/(sec^2x)$

But ${sec}^{2} x = \frac{1}{{cos}^{2} x}$ $" "sec^2x = 1/(cos^2x)$

Giving $({tan}^{2} x) ({cos}^{2} x)$ $" " (tan^2x)(cos^2x)$

But ${tan}^{2} x = \frac{{sin}^{2} x}{{cos}^{2} x}$ $tan^2x =(sin^2x)/(cos^2x)$

Giving $({sin}^{2} x) \frac{{cos}^{2} x}{{cos}^{2} x} = {sin}^{2} x$ $" " (sin^2x) (cos^2x)/(cos^2x) = sin^2x$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$ $color(red)("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")$
Clarification: By Tony B

Given: $\frac{{sec}^{2} x - 1}{{sec}^{2} x}$ $" " (sec^2x-1)/sec^2x$

But ${sec}^{2} x = 1 + {tan}^{2} x$ $Sec^2x = 1 +tan^2x$

Giving: $\frac{1 + {tan}^{2} x - 1}{{sec}^{2} x} = \frac{{tan}^{2} x}{{sec}^{2} x}$ $" "(1+tan^2x-1)/(sec^2x)" "=" " tan^2x/sec^2x$

But $tan x = \frac{sin x}{cos x} and sec x = \frac{1}{cos x}$ $" "tanx=sinx/cosx" and "secx= 1/cosx$

$" "sin^2x/(cancel(cos^2x)) xxcancel(cos^2x)" "=" "sin^2x$

Answer 3 · 2016-03-05T13:37:54

$=sin^2x$

Explanation:

$(sec^2x-1)/sec^2x$
$=cancel(sec^2x)/cancel(sec^2x)-1/sec^2x$
$=1-cos^2x$
$=sin^2x$

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