How do you prove #1+tan^2 (x) = sec^2 (x)#?

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Answer 1 · 2016-05-02T21:37:32

Using the following:

for #cos(x)!=0#, we have

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

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

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

#=1/cos^2(x)#

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

#=sec^2(x)#