Trigonometrical Ratios — Assertion-Reason Type Questions

Assertion (A): In the adjoining figure, tan A = $\dfrac{1}{\sqrt{3}}$. Then AC = 2AB.

Reason (R): In right angled ΔABC, AC² = AB² + BC²

1. Assertion (A) is true, Reason (R) is false.

2. Assertion (A) is false, Reason (R) is true.

3. Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Answer

Given, tan A = $\dfrac{1}{\sqrt{3}}$

As we know,

$\text{tan A }= \dfrac{\text{Perpendicular}}{\text{Base}}\\[1em] \Rightarrow \dfrac{BC}{AB} = \dfrac{1}{\sqrt{3}}$

Let BC = k and AB = $\sqrt{3}$ k

Since, ΔABC is a right angled triangle, using pythagoras theorem,

⇒ AC² = AB² + BC²

∴ Reason (R) is true.

Substituting values we get :

$\Rightarrow AC^2 = (\sqrt{3}k)^2 + k^2\\[1em] \Rightarrow AC^2 = 3k^2 + k^2\\[1em] \Rightarrow AC^2 = 4k^2\\[1em] \Rightarrow AC = \sqrt{4k^2}\\[1em] \Rightarrow AC = 2k \\[1em] \Rightarrow AC = 2k\dfrac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow AC = 2k \times \dfrac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow AC = \dfrac{2(\sqrt{3}k)}{\sqrt{3}} \\[1em] \Rightarrow AC = \dfrac{2AB}{\sqrt{3}}.$

∴ Assertion (A) is false.

∴ Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

Assertion (A): In adjoining triangle ABC, sinA cosA = $\dfrac{12}{25}$.

Reason (R): cos A = $\dfrac{1}{\text{sin A}}$.

1. Assertion (A) is true, Reason (R) is false.

2. Assertion (A) is false, Reason (R) is true.

3. Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Answer

In triangle ABC,

Using pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = 4² + 3²

⇒ AC² = 16 + 9

⇒ AC² = 25

⇒ AC = $\sqrt{25}$

⇒ AC = 5

By formula,

$\text{sin A} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}}\\[1em] = \dfrac{BC}{AC}\\[1em] = \dfrac{3}{5}.$

By formula,

$\text{cos A} = \dfrac{\text{Base}}{\text{Hypotenuse}}\\[1em] = \dfrac{AB}{AC}\\[1em] = \dfrac{4}{5}.$

Substituting values we get,

sin A cos A = $\dfrac{3}{5} \times \dfrac{4}{5} = \dfrac{3 \times 4}{5 \times 5} = \dfrac{12}{25}$.

∴ Assertion (A) is true.

By formula,

$\dfrac{1}{\text{sin A}}$ = cosec A ≠ cos A

∴ Reason (R) is false.

∴ Assertion (A) is true, Reason (R) is false.

Hence, option 1 is the correct option.

Assertion (A): If x = a cos θ + b sin θ and y = a cos θ - b sin θ, then x² + y² = a² + b²

Reason (R): cos²θ + sin²θ = 1.

1. Assertion (A) is true, Reason (R) is false.

2. Assertion (A) is false, Reason (R) is true.

3. Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Answer

Given, x = a cos θ + b sin θ and y = a cos θ - b sin θ

⇒ x² = (a cos θ + b sin θ)²

⇒ x² = a² cos² θ + b² sin² θ + 2 ab cos θ sin θ .....................(1)

Similarly,

⇒ y² = (a cos θ - b sin θ)²

⇒ y² = a² cos² θ + b² sin² θ - 2 a b cos θ sin θ .....................(2)

Adding equations (1) and (2), we get :

⇒ x² + y² = (a² cos² θ + b² sin² θ + 2ab cos θ sin θ) + (a² cos² θ + b² sin² θ - 2ab cos θ sin θ)

⇒ x² + y² = (a² cos² θ + a² cos² θ) + (b² sin² θ + b² sin² θ) + (2ab cos θ sin θ - 2ab cos θ sin θ)

⇒ x² + y² = 2a² cos² θ + 2 b² sin² θ

⇒ x² + y² = 2(a² cos² θ + b² sin² θ)

∴ Assertion (A) is false.

cos² θ + sin² θ = 1

This is a fundamental Pythagorean trigonometric identity and is always true for any real value of θ.

∴ Reason (R) is true.

∴ Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.