How do you verify $1-cos^2x=tan^2(-x)cos^2x$ ?

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Answer 1 · 2016-11-04T14:59:39

Verifying the equation of trigonometric identities is determined by applying some properties

$color(red)(tanx=sinx/cosx)$

$color(blue)(tan(-x)=-tanx)$

$color(purple)(cos^2x+sin^2x=1)$

Verifying the equality:
$tan^2(-x)cos^2x=(tan(-x))^2cos^2x$

$tan^2(-x)cos^2x=(color(blue)(-tan(x)))^2cos^2x$

$tan^2(-x)cos^2x=(-tan(x))^2cos^2x$

$tan^2(-x)cos^2x=tan^2xcos^2x$

$tan^2(-x)cos^2x=color(red)((sinx/cosx)^2cos^2x$

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

$tan^2(-x)cos^2x=color(violet)(sin^2x$

We have
$color(purple)(cos^2x+sin^2x=1)$
$rArrcolor(violet)(sin^2x=1-cos^2x$

Therefore,
$tan^2(-x)cos^2x=color(violet)(1-cos^2x$

How do you verify 1-cos^2x=tan^2(-x)cos^2x1-cos^2x=tan^2(-x)cos^2x?