How do you verify ${sec}^{4} x - {tan}^{4} x = {sec}^{2} x + {tan}^{2} x$ $Sec^4x-tan^4x=sec^2x+tan^2x$ ?

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How do you verify ${sec}^{4} x - {tan}^{4} x = {sec}^{2} x + {tan}^{2} x$ $Sec^4x-tan^4x=sec^2x+tan^2x$ ?

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Answer 1 · 2016-09-13T14:56:36

Prove trig expression

Explanation:

Transform the left side of the expression:
$L S = {sec}^{4} x - {tan}^{4} x = ({sec}^{2} x - {tan}^{2} x) ({sec}^{2} x + {tan}^{2} x)$ $LS = sec^4 x - tan^4 x = (sec^2 x - tan^2 x)(sec^2 x + tan^2 x)$ .
Since the first factor,
$({sec}^{2} x - {tan}^{2} x) = (\frac{1}{{cos}^{2} x} - \frac{{sin}^{2} x}{{cos}^{2} x}) =$ $(sec^2 x - tan^2 x) = (1/(cos^2 x) - (sin^2 x)/(cos^2 x)) =$
$= \frac{1 - {sin}^{2} x}{{cos}^{2} x} = \frac{{cos}^{2} x}{{cos}^{2} x} = 1$ $= (1 - sin^2 x)/(cos^2 x) = (cos^2 x)/(cos^2 x) = 1$
There for, the left side becomes;
$L S = ({sec}^{2} x + {tan}^{2} x)$ $LS = (sec^2 x + tan^2 x)$ , and it equals the right side.

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How do you verify ${sec}^{4} x - {tan}^{4} x = {sec}^{2} x + {tan}^{2} x$ $Sec^4x-tan^4x=sec^2x+tan^2x$ ?

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How do you verify sec4x−tan4x=sec2x+tan2xSec^4x-tan^4x=sec^2x+tan^2x?

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How do you verify ${sec}^{4} x - {tan}^{4} x = {sec}^{2} x + {tan}^{2} x$ $Sec^4x-tan^4x=sec^2x+tan^2x$ ?