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Let x = 3 sec theta!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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Find the sqrt [x^2 -9]/x^2?
asked Apr 13, 2013 in TRIGONOMETRY by angel12 Scholar

1 Answer

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x = 3secθ

Take square to each side

x2 = 9sec2θ

Recall : Trogonometry formulas sec2θ - tan2θ = 1

Substitute sec2θ = 1 + tan2θ in the x2 = 9sec2θ

x2 = 9(1 + tan2θ)

Recall : Distributive property a(b + c) = ab + ac

x2 = 9 + 9tan2θ

Subtract 9 from each side

x2 - 9 = 9tan2θ

Take square root to each side

√[x2 - 9] = 3tanθ

√[x2 - 9] / x2 = 3tanθ / 9sec2θ

√[x2 - 9] / x2 = tanθ / 3sec2θ

√[x2 - 9] / x^2 = tanθ*cos^2θ / 3

Simplify

√[x2 - 9] / x2 = (sinθ / cosθ)*cos2θ / 3

√[x2 - 9] / x2 = (sinθ * cosθ) / 3

√[x2 - 9] / x2 = (sin2θ) / 6.

answered Apr 13, 2013 by diane Scholar

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