Question

Prove 2/sqrt3cosx+sinx=sec(pi/6-x)

Answer 1

Thus L.H.S = R.H.S that is 2/√3cosx + sinx  = sec(Π/6-x) is proved

We have to prove that

2/√3cosx + sinx  = sec(Π/6-x)

To prove this we will solve the right-hand side of the equation which is

R.H.S = sec(Π/6-x)

          = 1/cos(Π/6-x)

[As secƟ = 1/cosƟ)

           = 1/[cos Π/6cosx + sin Π/6sinx]

[As cos (X-Y) = cosXcosY + sinXsinY , which is a trigonometry identity where X = Π/6 and Y = x]

           = 1/[√3/2cosx + 1/2sinx]

            = 1/(√3cosx + sinx]/2

            = 2/√3cosx + sinx

    R.H.S = L.H.S

Hence 2/√3cosx + sinx  = sec(Π/6-x) is proved

Learn more about trigonometry here : brainly.com/question/7331447

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