How do you prove ${sec}^{2} x - {tan}^{2} x = {cos}^{2} x + {sin}^{2} x$ $sec^2x-tan^2x =cos^2x+sin^2x$ ?

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Answer 1 · 2018-05-18T02:01:27

See below

Using:
$1 + {sec}^{2} x = {tan}^{2} x$ $1+sec^2x=tan^2x$
$1 = {sin}^{2} x + {cos}^{2} x$ $1=sin^2x+cos^2x$

Start:
${sec}^{2} x - {tan}^{2} x = {cos}^{2} x + {sin}^{2} x$ $sec^2x-tan^2x =cos^2x+sin^2x$

$(1 + {tan}^{2} x) - {tan}^{2} x = {cos}^{2} x + {sin}^{2} x$ $(1+tan^2x)-tan^2x=cos^2x+sin^2x$

$1+cancel(tan^2x)cancel(-tan^2x)=cos^2x+sin^2x$

$1=cos^2x+sin^2x$

$cos^2x+sin^2x=cos^2x+sin^2x$

How do you prove sec^2x-tan^2x =cos^2x+sin^2x sec2x−tan2x=cos2x+sin2x sec^2x-tan^2x =cos^2x+sin^2x ?