How do I prove #sec^2x/(2+tan^2x) = 1/(1+cos^2x)#?

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Answer 1 · 2018-04-28T19:27:13

See below

Using:
#cosx=1/secx#
#1+tan^2x=sec^2x#

START:

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

#sec^2x/(2+tan^2x)=1/(1+1/sec^2x)#

#sec^2x/(2+tan^2x)=1/(sec^2x/sec^2x+1/sec^2x)#

#sec^2x/(2+tan^2x)=1/((sec^2x+1)/sec^2x)#

#sec^2x/(2+tan^2x)=sec^2x/(sec^2x+1)#

#sec^2x/(2+tan^2x)=sec^2x/((1+tan^2x)+1)#

#sec^2x/(2+tan^2x)=sec^2x/(2+tan^2x) quad sqrt#

END

Answer 2 · 2018-04-28T19:55:31

Kindly go through a Proof in the Explanation.

Knowing that, #sec^2x=tan^2x+1#,

we have, #sec^2x/(2+tan^2x)#,

#=sec^2x/{2+(sec^2x-1)}#,

#=sec^2x/(sec^2x+1)#,

#=sec^2x/(sec^2x+1)xxcos^2x/cos^2x#,

#=(sec^2xcos^2x)/(sec^2xcos^2x+cos^2x)#,

#=1/(1+cos^2x)#, as desired!