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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