# 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)

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