Prove that #(1+tan^2x)cos^2x=1#?

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Answer 1 · 2018-04-19T18:57:35

See below

We will use the following identities

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

#sec^nx=1/cos^nx#

#tan^nx=sin^nx/cos^nx#

Proof

We start by manipulating the first identity

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

#rArr cos^2/cos^2x+sin^2x/cos^2x=1/cos^2x#

#rArr1+tan^2x=sec^2x#

So

#(1+tan^2x)cos^2x=sec^2xcos^2x=1/cos^2xcos^2x=1# #color(white)(aaaa)square#