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

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Answer 1 · 2018-05-18T02:01:27

See below

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

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

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

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

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

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

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