Trigonometrical Ratios of Standard Angles — Multiple Choice Questions

The value of $\dfrac{\text{tan 30°}}{\text{cot 60°}}$ is

1. $\dfrac{1}{\sqrt{2}}$

2. $\dfrac{1}{\sqrt{3}}$

3. $\sqrt{3}$

4. 1

Answer

Solving,

$\Rightarrow \dfrac{\dfrac{1}{\sqrt{3}}}{\dfrac{1}{\sqrt{3}}} \\[1em] \Rightarrow 1.$

Hence, Option 4 is the correct option.

The value of (sin 45° + cos 45°) is

1. $\dfrac{1}{\sqrt{2}}$

2. $\sqrt{2}$

3. $\dfrac{\sqrt{3}}{2}$

4. 1

Answer

Solving,

$\text{sin 45° + cos 45°} = \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}} \\[1em] = \dfrac{2}{\sqrt{2}} \\[1em] = \sqrt{2}.$

Hence, Option 2 is the correct option.

The value of tan² 30° - 4 sin² 45° is

1. 1

2. $\dfrac{7}{3}$

3. $-\dfrac{5}{3}$

4. $-\dfrac{11}{3}$

Answer

Solving,

$\Rightarrow \text{tan}^2 30° - 4\text{sin}^2 45° = \Big(\dfrac{1}{\sqrt{3}}\Big)^2 - 4 \times \Big(\dfrac{1}{\sqrt{2}}\Big)^2 \\[1em] = \dfrac{1}{3} - 4 \times \dfrac{1}{2} \\[1em] = \dfrac{1}{3} - 2 \\[1em] = \dfrac{1 - 6}{3} \\[1em] = -\dfrac{5}{3}.$

Hence, Option 3 is the correct option.

If A = 30°, then the value of 2 sin A cos A is

1. $\dfrac{1}{\sqrt{2}}$

2. $\dfrac{\sqrt{3}}{2}$

3. $\dfrac{1}{2}$

4. 1

Answer

Solving,

$\Rightarrow \text{2 sin A cos A} = \text{2 sin 30° cos 30°} \\[1em] = 2 \times \dfrac{1}{2} \times \dfrac{\sqrt{3}}{2} \\[1em] = \dfrac{\sqrt{3}}{{2}}.$

Hence, Option 2 is the correct option.

The value of (sin 30° + cos 30°) - (sin 60° + cos 60°) is

1. -1

2. 0

3. 1

4. 2

Answer

Solving,

$\text{(sin 30° + cos 30°) - (sin 60° + cos 60°)} = \Big(\dfrac{1}{2} + \dfrac{\sqrt{3}}{{2}}\Big) - \Big(\dfrac{\sqrt{3}}{2} + \dfrac{1}{2}\Big) \\[1em] = \dfrac{1}{2} + \dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{3}}{2} - \dfrac{1}{2} \\[1em] = 0.$

Hence, Option 2 is the correct option.

The value of $\sqrt{3}$ cosec 60° - sec 60° is

1. 0

2. 1

3. 2

4. -1

Answer

Solving,

$\Rightarrow \sqrt{3}\text{ cosec 60° - sec 60°} = \sqrt{3} \times \dfrac{2}{\sqrt{3}} - 2 \\[1em] = 2 - 2 \\[1em] = 0.$

Hence, Option 1 is the correct option.

The value of $\dfrac{1}{\text{sin 30°}} - \dfrac{\sqrt{3}}{\text{cos 30°}}$ is

1. 2

2. 1

3. $\dfrac{1}{2}$

4. 0

Answer

Solving,

$\Rightarrow \dfrac{1}{\text{sin 30°}} - \dfrac{\sqrt{3}}{\text{cos 30°}} = \dfrac{1}{\dfrac{1}{2}} - \dfrac{\sqrt{3}}{\dfrac{\sqrt{3}}{2}} \\[1em] = 2 - 2 \\[1em] = 0.$

Hence, Option 4 is the correct option.

If tan A = $\sqrt{3}$, then the value of cosec A is

1. $\dfrac{1}{2}$

2. 2

3. $\dfrac{2}{\sqrt{3}}$

4. $\dfrac{\sqrt{3}}{2}$

Answer

Given,

⇒ tan A = $\sqrt{3}$

⇒ tan A = tan 60°

⇒ A = 60°.

⇒ cosec A = cosec 60° = $\dfrac{2}{\sqrt{3}}$.

Hence, Option 3 is the correct option.

If sec θ. sin θ = 0, then the value of cos θ is

1. 0

2. $\dfrac{1}{\sqrt{2}}$

3. $\dfrac{1}{2}$

4. 1

Answer

Given,

$\Rightarrow \text{sec θ. sin θ = 0} \\[1em] \Rightarrow \dfrac{1}{\text{cos θ}} \times \text{sin θ} = 0 \\[1em] \Rightarrow \text{tan θ} = 0 \\[1em] \Rightarrow θ = 0°.$

⇒ cos θ = cos 0° = 1.

Hence, Option 4 is the correct option.

If sin α = $\dfrac{1}{2},$ then the value of 3 cos α - 4 cos³ α is

1. -1

2. 0

3. 1

4. 2

Answer

Given,

⇒ sin α = $\dfrac{1}{2}$

⇒ sin α = sin 30°

⇒ α = 30°.

$\text{3 cos α - 4 cos}^3 α = 3\text{ cos 30° - 4 cos}^3 30° \\[1em] = 3 \times \dfrac{\sqrt{3}}{2} - 4 \times \Big(\dfrac{\sqrt{3}}{2}\Big)^3 \\[1em] = \dfrac{3\sqrt{3}}{2} - 4 \times \dfrac{3\sqrt{3}}{8} \\[1em] = \dfrac{3\sqrt{3}}{2} - \dfrac{3\sqrt{3}}{2} \\[1em] = 0.$

Hence, Option 2 is the correct option.

The value of $\dfrac{\text{1 - tan}^2 45°}{\text{1 + tan}^2 45°}$ is equal to

1. tan 60°

2. tan 30°

3. sin 45°

4. tan 0°

Answer

Solving,

$\Rightarrow \dfrac{\text{1 - tan}^2 45°}{\text{1 + tan}^2 45°} = \dfrac{1 - 1}{1 + 1} \\[1em] = 0 \\[1em] = \text{tan 0°}.$

Hence, Option 4 is the correct option.

If sin α = $\dfrac{1}{2}$ and cos β = $\dfrac{1}{2}$, then the value of (α + β) is

1. 0°

2. 30°

3. 60°

4. 90°

Answer

Given,

⇒ sin α = $\dfrac{1}{2}$

⇒ sin α = sin 30°

⇒ α = 30°.

Also,

⇒ cos β = $\dfrac{1}{2}$

⇒ cos β = cos 60°

⇒ β = 60°.

(α + β) = 30° + 60° = 90°.

Hence, Option 4 is the correct option.

If △ABC is right angled at C, then the value of cos (A + B) is

1. 0

2. 1

3. $\dfrac{1}{2}$

4. $\dfrac{\sqrt{3}}{2}$

Answer

In △ABC,

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A + ∠B + 90° = 180°

⇒ ∠A + ∠B = 90°.

⇒ cos (A + B) = cos 90° = 0.

Hence, Option 1 is the correct option.

In the adjoining figure, ABC is a right triangle right angled at B. If AB = 10 cm and ∠C = 30°, then the length of the side BC is

1. $\dfrac{10}{\sqrt{3}}$ cm

2. $10\sqrt{3}$ cm

3. 20 cm

4. 5 cm

Answer

By formula,

tan C = $\dfrac{\text{Perpendicular}}{\text{Base}}$

$\Rightarrow \text{tan 30°} = \dfrac{AB}{BC} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{10}{BC} \\[1em] \Rightarrow BC = 10\sqrt{3} \text{ cm}.$

Hence, Option 2 is the correct option.

In the adjoining figure, PQR is a right triangle right angled at Q. If PQ = 4 cm and PR = 8 cm then ∠P is equal to

1. 60°

2. 45°

3. 30°

4. 15°

Answer

By formula,

cos P = $\dfrac{\text{Base}}{\text{Hypotenuse}}$

$\Rightarrow \text{cos P} = \dfrac{PQ}{PR} \\[1em] \Rightarrow \text{cos P} = \dfrac{4}{8} \\[1em] \Rightarrow \text{cos P} = \dfrac{1}{2} \\[1em] \Rightarrow \text{cos P} = \text{cos 60°} \\[1em] \Rightarrow P = 60°.$

Hence, Option 1 is the correct option.

Consider the following two statements.

Statement 1: sin 18° - cos 72° = 0.

Statement 2: sin θ = cos (90° - θ).

Which of the following is valid?

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and Statement 2 is false.

4. Statement 1 is false, and Statement 2 is true.

Answer

Given,

sin 18° - cos 72° = 0

Solving L.H.S.,

⇒ sin (90° - 72°) - cos 72°

⇒ cos 72° - cos 72°

⇒ 0.

Since, L.H.S. = R.H.S. = 0.

∴ Statement 1 is true.

The statement sin θ = cos (90° - θ) is true.

This is a fundamental trigonometric identity, often referred to as the cofunction identity.

∴ Statement 2 is true.

∴ Both the statements are true.

Hence, option 1 is correct option.