How do you determine graphically and analytically whether $y_{1} = {sec}^{2} x - 1$ $y_1=sec^2x-1$ is equivalent to $y_{2} = {tan}^{2} x$ $y_2=tan^2x$ ?

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How do you determine graphically and analytically whether $y_{1} = {sec}^{2} x - 1$ $y_1=sec^2x-1$ is equivalent to $y_{2} = {tan}^{2} x$ $y_2=tan^2x$ ?

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Answer 1 · 2018-02-24T15:49:55

See below.

Explanation:

Identities:

1) $88 {sin}^{2} x + {cos}^{2} x = 1$ $color(white)(88)color(red)bb(sin^2x+cos^2x=1)$

2) $88 {sec}^{2} x = \frac{1}{{cos}^{2} x}$ $color(white)(88)color(red)bb(sec^2x=1/cos^2x)$

3) $88 {tan}^{2} x = \frac{{sin}^{2} x}{{cos}^{2} x}$ $color(white)(88)color(red)bb(tan^2x=sin^2x/cos^2x)$

$y_{1} = y_{2} \Rightarrow {sec}^{2} x - 1 = {tan}^{2} x$ $y_1=y_2=>sec^2x-1=tan^2x$

LHS

By identity 2.

$\frac{1}{{cos}^{2} x} - 1$ $1/cos^2x-1$

Add:

$\frac{1 - {cos}^{2} x}{{cos}^{2} x}$ $(1-cos^2x)/cos^2x$

From identity 1:

$1 - {cos}^{2} x = {sin}^{2} x$ $1-cos^2x=sin^2x$

Hence:

$\frac{{sin}^{2} x}{{cos}^{2} x}$ $sin^2x/cos^2x$

By identity 3:

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

So:

$L H S \equiv R H S$ $LHS-=RHS$

This is an identity, so they are equivalent.

Graphically:

Plot both $bb(sec^2x-1)$ and $bb(tan^2x)$ . As you can see their graphs are identical, as we would expect.

$bb(y=sec^2x-1)$

graph{y=(tan(x))^2 [-11.25, 11.25, -5.625, 5.625]}

$bb(y=tan^2x)$
graph{y=(sec(x))^2-1 [-11.25, 11.25, -5.625, 5.625]}

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How do you determine graphically and analytically whether $y_{1} = {sec}^{2} x - 1$ $y_1=sec^2x-1$ is equivalent to $y_{2} = {tan}^{2} x$ $y_2=tan^2x$ ?

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How do you determine graphically and analytically whether y_1=sec^2x-1y1=sec2x−1y_1=sec^2x-1 is equivalent to y_2=tan^2xy2=tan2xy_2=tan^2x?

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How do you determine graphically and analytically whether $y_{1} = {sec}^{2} x - 1$ $y_1=sec^2x-1$ is equivalent to $y_{2} = {tan}^{2} x$ $y_2=tan^2x$ ?