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

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Answer 1 · 2017-03-09T16:51:52

Please see below.

#cos^2+tan^2cos^2x#

= #cos^2+sin^2x/cos^2x xxcos^2x#

= #cos^2+sin^2x/cancel(cos^2x) xxcancel(cos^2x)#

= #cos^2+sin^2x#

= #1#

Answer 2 · 2017-03-09T16:53:02

Factor out #cos^2x#.

#cos^2x(1 + tan^2x) = 1#

Use #tan^2x + 1 = sec^2x#.

#cos^2x(sec^2x) = 1#

Use the identity #secx= 1/cosx#.

#cos^2x(1/cos^2x) =1#

#1 = 1#

This identity is true.

Hopefully this helps!