Question

Question #f2c09

Answer 1

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)`

Answer 2

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

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

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

`=1/cosx`

`= secx=RHS`

Proved

Answer 3

look at picture

Explanation:

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