Question #f2c09

3 Answers
Jun 3, 2017

Recall the following identities:

  • sin2x = sin(x+x) = sinxcosx + cosxsinx

= 2sinxcosx

  • cos2x = cos(x+x) = cosxcosx - sinxsinx

= cos^2x - sin^2x

  • sin^2x + cos^2x = 1

This gives:

(2(sinx + cosx))/(2sinxcosx + cos^2x - sin^2x + 1)

= (2(sinx + cosx))/(2sinxcosx + cos^2x - cancel(sin^2x) + cancel(sin^2x) + cos^2x)

= (cancel(2)(sinx + cosx))/(cancel(2)(sinxcosx + cos^2x))

= cancel(sinx + cosx)/(cosxcancel((sinx + cosx)))

= 1/cosx -= color(blue)(secx)

Jun 3, 2017

LHS=(2(cosx+sinx))/(sin2x+cos2x+1)

=(2(cosx+sinx))/(2sinxcos+2cos^2x)

=(2(cosx+sinx))/(2cos(sinx+cosx))

=1/cosx

= secx=RHS

Proved

Jun 3, 2017

look at picture

Explanation:

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