How do you prove the following problem step by step and which 4 identities are used to prove?

2/(sqrt3 cosx+sinx)=sec(pi/6-x)23cosx+sinx=sec(π6x)

3 Answers
Apr 8, 2018

Two identities needed

Explanation:

Well let's manipulate the right side to match the left, and name identities as encountered:

2/(sqrt3cosx+sinx)= sec(pi/6-x)23cosx+sinx=sec(π6x)

Reciprocal identity: secx= 1/cosxsecx=1cosx
2/(sqrt3cosx+sinx)= 1/cos(pi/6-x)23cosx+sinx=1cos(π6x)

Cosine difference identity: cos(x-y)=cosxcosy+sinxsinycos(xy)=cosxcosy+sinxsiny
2/(sqrt3cosx+sinx)= 1/(cos(pi/6)cosx+sin(pi/6)sinx)23cosx+sinx=1cos(π6)cosx+sin(π6)sinx

Simplify:
2/(sqrt3cosx+sinx)= 1/((sqrt3cosx)/2+sinx/2)23cosx+sinx=13cosx2+sinx2

2/(sqrt3cosx+sinx)= 1/((sqrt3cosx+sinx)/2)23cosx+sinx=13cosx+sinx2

Reciprocate:
2/(sqrt3cosx+sinx)= 2/(sqrt3cosx+sinx)23cosx+sinx=23cosx+sinx

Apr 8, 2018

Image reference...

Explanation:

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Apr 8, 2018

Relations used

1.cos(pi/6)=sqrt3/2cos(π6)=32

2.sin(pi/6)=1/2sin(π6)=12

3.cosAcosB+sinAsinB=cos(A-B)cosAcosB+sinAsinB=cos(AB)

4.1/costheta=sectheta1cosθ=secθ

LHS=2/(sqrt3 cosx+sinx)LHS=23cosx+sinx

=(2/2)/(sqrt3/2 cosx+1/2sinx)=2232cosx+12sinx

=1/(cos(pi/6) cosx+sin(pi/6)sinx)=1cos(π6)cosx+sin(π6)sinx

=1/(cos(pi/6-x))=1cos(π6x)

=sec(pi/6-x)=sec(π6x)