How do you derive #sec^2x=tan^x+1# from #sin^2x+cos^2x=1#?

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Answer 1 · 2016-05-31T18:34:45

See derivation.

#color(red)("Assumption: question should read as "sec^2x=tan^2x+1)#

Given:#" "sin^2x+cos^2x=1#

Multiply both sides by #1/cos^2x#

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

'...............................................
But
#sin^2x/cos^2x=tan^2x#

#1/cos^2x=sec^2x#

#cos^2x/cos^2x=1#
'...................................................
Giving:

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

Changing the order

#color(green)(sec^2x=tan^2x+1)#