Question

Prove that
`2/√(3) cosx + sin x= sec(\pi /6-x)`

Answer 1

sec(π/6 - x) = R.H.S = 2/(√3cosx + sinx) = L.H.S.

Explanation:

sec(π/6 - x) = 1/cos(π/6 - x)

Using compound angle formula,

cos(A - B) = cosAcosB + sinAsinB where A = π/6 and B = x.

So, cos(π/6 - x) = cos(π/6)cosx + sin(π/6)sinx , cosπ/6 = √3/2 and sinπ/6 = 1/2

cos(π/6 - x) = cosπ/6cosx + sinπ/6sinx = (√3/2)cosx + (1/2)sinx

sec(π/6 - x) = 1/cos(π/6 - x)

= 1/(√3/2)cosx + (1/2)sinx = 2/(√3cosx + sinx)

= L.H.S

So, sec(π/6 - x) = R.H.S = 2/(√3cosx + sinx) = L.H.S