Worksheet on Elimination of Unknown Angle(s) Using Trigonometric Identities

In Worksheet on elimination of unknown angle(s) using Trigonometric identities we will prove various types of practice questions on Trigonometric identities.

Here you will get 11 different types of elimination of unknown angle using Trigonometric identities questions with some selected questions hints.

1. Eliminate θ (theta) in each of the following:

(i) x = a sec θ, y = b tan θ

(ii) a sin θ = p, b tan θ = q

(iii) sin θ + cos θ = m, tan θ + cot θ = n

(iv) sin θ – cos θ = m, sec θ - csc θ = b

2. If sin θ + cos θ = m and sec θ + csc θ = n, then prove that

n(m² – 1) = 2m.

Hint: n = sec θ + csc θ

    ⟹ n = \(\frac{1}{cos θ}\) + \(\frac{1}{sin θ}\) 

    ⟹ n = \(\frac{sin θ + cos θ}{sin θ cos θ}\) 

    ⟹ n = \(\frac{m}{sin θ cos θ}\) 

    ⟹ sin θ cos θ = \(\frac{m}{n}\) ......... (i) 

Now, m² – 1 = (sin θ + cos θ)² - 1 

                   = (sin² θ + sin² θ + 2 sin θ cos θ) - 1 

                   = 1 + 2 sin θ cos θ - 1 

                   = 2 sin θ cos θ

                   = 2\(\frac{m}{n}\), From (i)

3. If l₁ cos θ + m₁ sin θ + n₁ = 0 and l₂ cos θ + m₂ sin θ + n₂ = 0 then prove that

(m₁n₂ – n₁m₂)² + (n₁l₂ – n₂l₁)² = (l₁m₂ – l₂m₁)²

4. If a sin² ϕ + b cos² ϕ = c and p sin² ϕ + q cos² ϕ = r then prove that

(b – c)(r – p) = (c – a)(q – r).

Hint: \(\frac{b - c}{c - a}\) = \(\frac{b - (a sin^{2} ϕ + b cos^{2} ϕ)}{(a sin^{2} ϕ + b cos^{2} ϕ) - a}\)

              = \(\frac{(b - a) sin^{2} ϕ}{(b - a) cos^{2} ϕ}\)

                 = tan² ϕ.

Similarly, \(\frac{q - r}{r - p}\) = \(\frac{q - (p sin^{2} ϕ + q cos^{2} ϕ)}{(p sin^{2} ϕ + q cos^{2} ϕ) - p}\)

                      = \(\frac{(q - p) sin^{2} ϕ}{(q - p) cos^{2} ϕ}\)

                      = tan² ϕ.

Therefore, \(\frac{b - c}{c - a}\) = \(\frac{q - r}{r - p}\).

5. If a sec θ + b tan θ + c = 0 and a’ sec θ + b’ tan θ + c’ = 0 then prove that

(bc’ – b’c)² – (ca’ – ac’)² = (ab’ – a’b)².

6. If \(\frac{x}{a cos θ}\) = \(\frac{y}{b sin θ}\) and \(\frac{ax}{cos θ}\) - \(\frac{by}{sin θ}\) = a² – b², prove that

\(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1.

Hint: \(\frac{x}{cos θ}\) ∙ b - \(\frac{y}{sin θ}\) ∙ a + 0 = 0 and \(\frac{x}{cos θ}\) ∙ a - \(\frac{y}{sin θ}\) ∙ b - (a² - b²) = 0.

By cross multiplication, \(\frac{\frac{x}{cos θ}}{a(a^{2} - b^{2})}\) = \(\frac{\frac{y}{sin θ}}{b(a^{2} - b^{2})}\) = \(\frac{1}{(a^{2} - b^{2})}\)

⟹ \(\frac{x}{a}\) = cos θ,  \(\frac{y}{b}\) = sin θ. Square these and add.

7. If tan A + sin A = m and tan A - sin A = n then prove that

m² – n² = 4 \(\sqrt{mn}\).

8. If x sin³ A + y cos³ A = sin A ∙ cos A and x sin A – y cos A = 0 then prove that

x² + y² = 1.

Hint: x sin A - y cos A = 0 

⟹ tan A = \(\frac{y}{x}\)

Again, x ∙ \(\frac{sin^{2} A}{cos A}\) + y ∙ \(\frac{cos^{2} A}{sin A}\) = 1

⟹ x ∙ \(\frac{y}{x}\) sin A + y ∙ \(\frac{x}{y}\) cos A = 1

⟹ x cos A + y sin A = 1

Now, (x sin A - y cos A)² + (x cos A + y sin A)² = 0² + 1²

9. If csc β – sin β = m³; sec β – cos β = n³ then prove that,

m²n²(m² + n²) = 1.

10. If a = r cos θ cos β, b = r cos θ sin β and c = r sin θ then prove that,

a² + b² + c² = r².

11. If p = a sec A cos B, q = b sec A sin B and r = c tan A then prove that, 

\(\frac{p^{2}}{a^{2}}\) + \(\frac{q^{2}}{b^{2}}\) - \(\frac{r^{2}}{c^{2}}\) = 1.

Answers

1. (i) \(\frac{x^{2}}{a^{2}}\) - \(\frac{y^{2}}{b^{2}}\) = 1.

(ii) \(\frac{a^{2}}{p^{2}}\) - \(\frac{b^{2}}{q^{2}}\) = 1.

(iii) n(m² – 1) = 2

(iv) b(1 – a²) = 2a

10th Grade Math

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