Question #dcdc4

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Answer 1 · 2017-08-01T19:59:08

Start with:

#cos^2(theta) + sin^2(theta) = 1#

Divide both sides of the equation by #cos^2(theta)#

#1 + sin^2(theta)/cos^2(theta) = 1/cos^2(theta)#

Substitute #tan^2(theta)# for #sin^2(theta)/cos^2(theta)#:

#1 + tan^2(theta) = 1/cos^2(theta)#

Substitute #sec^2(theta)# for #1/cos^2(theta)#:

#1 + tan^2(theta) = sec^2(theta)# Q.E.D.

Answer 2 · 2017-08-01T20:02:09

See the proof below

We need

#tantheta=sintheta/costheta#

#sectheta=1/costheta#

We start with

#sin^2theta+cos^2theta=1#

Dividing throughout by #cos^2theta#

#sin^2theta/cos^2theta+cos^2theta/cos^2theta=1/cos^2theta#

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

#QED#