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1 year ago

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

Mathematics

1 answer:

dlinn [17]1 year ago

5 0

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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