Elimination of Unknown Angles

Problems on elimination of unknown angles using trigonometric identities.

1. If x = tan θ + sin θ and y = tan θ - sin θ, prove that x² – y² = 4\(\sqrt{xy}\).

Solution:

Given that

x = tan θ + sin θ ……………………. (i)

and

y = tan θ - sin θ ……………………. (ii)

Adding (i) and (ii), we get

x + y = 2 tan θ ……………………. (iii)

⟹ tan θ = \(\frac{x + y}{2}\) ……………………. (iv)

Subtracting (ii) from (i), we get,

x - y = 2 sin θ ……………………. (v)

Now, dividing (iii) by (v) we get,

\(\frac{x + y}{x - y}\) = \(\frac{2 tan θ}{2 sin θ}\)

       = \(\frac{tan θ}{sin θ}\)

       = \(\frac{\frac{sin θ}{cos θ}}{sin θ}\)

       = \(\frac{sin θ}{cos θ}\) ∙ \(\frac{1}{sin θ}\)

       = \(\frac{1}{cos θ}\)

       = sec θ

Therefore, sec θ = \(\frac{x + y}{x - y}\) ……………………. (vi)

We know that the Pythagorean identity, sec\(^{2}\) θ - tan\(^{2}\) θ = 1.

Now from (iv) and (vi) we get,

\((\frac{x + y}{x - y})^{2}\) - \((\frac{x + y}{2})^{2}\) = 1

Taking common (x + y)\(^{2}\) we get,

⟹ (x + y)\(^{2}\) ∙ {\(\frac{1}{(x - y)^{2}} - \frac{1}{4}\)} = 1

⟹ (x + y)\(^{2}\) ∙ \(\frac{4 – (x – y)^{2}}{4(x – y)^{2}}\)= 1

⟹ (x + y)\(^{2}\) ∙ {4 – (x – y)\(^{2}\)} = 4(x – y)\(^{2}\)

⟹ 4(x + y)\(^{2}\) - (x + y)\(^{2}\) ∙ (x – y)\(^{2}\) = 4(x – y)\(^{2}\)

⟹ 4(x + y)\(^{2}\) - 4(x – y)\(^{2}\) = (x + y)\(^{2}\) ∙ (x – y)\(^{2}\)

⟹ 4(x\(^{2}\) + y\(^{2}\) + 2xy - x\(^{2}\) - y\(^{2}\) + 2xy) = \((x^{2} + y^{2})^{2}\)

⟹ 4 ∙ 4xy = \((x^{2} + y^{2})^{2}\)

⟹ 16xy = \((x^{2} + y^{2})^{2}\)

⟹ 4\(\sqrt{xy}\) = \(x^{2} + y^{2}\)

Therefore, \(x^{2} + y^{2}\) = 4\(\sqrt{xy}\). (Proved)

2. If a = r cos θ ∙ sin  β, b = r cos θ ∙ cos  β and c = r sin θ then prove that a\(^{2}\) + b\(^{2}\) + c\(^{2}\) = r\(^{2}\).

Solution:

a\(^{2}\) + b\(^{2}\) + c\(^{2}\) = r\(^{2}\) cos\(^{2}\) θ ∙ sin\(^{2}\) β + r\(^{2}\)  cos\(^{2}\) θ ∙ cos\(^{2}\)  β + r\(^{2}\) sin\(^{2}\) θ

                   = r\(^{2}\) cos\(^{2}\) θ(sin\(^{2}\) β + cos\(^{2}\) β) + r\(^{2}\) sin\(^{2}\) θ

                   = r\(^{2}\) cos\(^{2}\) θ ∙ (1) + r\(^{2}\) sin\(^{2}\) θ, [since We know that the Pythagorean identity, sin\(^{2}\) θ + cos\(^{2}\) θ = 1.]

                   = r\(^{2}\) cos\(^{2}\) θ + r\(^{2}\) sin\(^{2}\) θ

                   = r\(^{2}\) (cos\(^{2}\) θ + sin\(^{2}\) θ)

                   = r\(^{2}\) ∙ (1), [since, sin\(^{2}\) θ + cos\(^{2}\) θ = 1]

                   = r\(^{2}\)

Therefore, a\(^{2}\) + b\(^{2}\) + c\(^{2}\) = r\(^{2}\). (proved)

10th Grade Math

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