Question #ac9e3

Question

1386 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-18T07:21:52

See the proof below

We use

$tanx=sinx/cosx$

$sin^2x+cos^2x=1$

LHS, $tan^2x/(secx+1)=(sin^2x/cos^2x)/(1/cosx+1)$

$=sin^2x/(cos^2x+cosx)$

$=(1-cos^2x)/(cosx(1+cosx))$

$=(cancel(1+cosx)(1-cosx))/(cosxcancel(1+cosx))$

$=(1-cosx)/cosx$

$=RHS$

$Q.E.D$

Answer 2 · 2016-12-18T07:26:25

$LHS=tan^2x/(secx+1)$

$=sin^2x/(cos^2x(secx+1))$

$=(1-cos^2x)/(cosx*cosx(secx+1))$

$=((1-cosx)(1+cosx))/(cosx(cosx*secx+cosx))$

$=((1-cosx)cancel((1+cosx)))/(cosxcancel((1+cosx)))$

$=(1-cosx)/cosx=RHS$

Proved