How do I prove sec^2x/(2+tan^2x) = 1/(1+cos^2x)sec2x2+tan2x=11+cos2x?

2 Answers
Apr 28, 2018

See below

Explanation:

Using:
cosx=1/secxcosx=1secx
1+tan^2x=sec^2x1+tan2x=sec2x

START:

sec^2x/(2+tan^2x)=1/(1+cos^2x)sec2x2+tan2x=11+cos2x

sec^2x/(2+tan^2x)=1/(1+1/sec^2x)sec2x2+tan2x=11+1sec2x

sec^2x/(2+tan^2x)=1/(sec^2x/sec^2x+1/sec^2x)sec2x2+tan2x=1sec2xsec2x+1sec2x

sec^2x/(2+tan^2x)=1/((sec^2x+1)/sec^2x)sec2x2+tan2x=1sec2x+1sec2x

sec^2x/(2+tan^2x)=sec^2x/(sec^2x+1)sec2x2+tan2x=sec2xsec2x+1

sec^2x/(2+tan^2x)=sec^2x/((1+tan^2x)+1)sec2x2+tan2x=sec2x(1+tan2x)+1

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

END

Apr 28, 2018

Kindly go through a Proof in the Explanation.

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!