Trigonometrical Ratios of Standard Angles — Assertion-Reason Type Questions

Assertion (A): tan 30° + sec 30° = cot 30°.

Reason (R): sec θ = $\dfrac{\text{cosec θ}}{\text{cot θ}}$

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 30° + sec 30° = cot 30°.

Solving, L.H.S.

⇒ tan 30° + sec 30°

$\Rightarrow \dfrac{1}{\sqrt{3}} + \dfrac{2}{\sqrt{3}}\\[1em] \Rightarrow \dfrac{1 + 2}{\sqrt{3}}\\[1em] \Rightarrow \dfrac{3}{\sqrt{3}}\\[1em] \Rightarrow \sqrt{3}$

Solving, R.H.S. = cot 30° = $\sqrt{3}$

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

∴ Assertion (A) is true.

According to reason (R) : sec θ = $\dfrac{\text{cosec θ}}{\text{cot θ}}$

Solving R.H.S.,

$\Rightarrow \dfrac{\text{cosec θ}}{\text{cot θ}}\\[1em] \Rightarrow \dfrac{\dfrac{1}{\text{sin θ}}}{\dfrac{\text{cos θ}}{\text{sin θ}}}\\[1em] \Rightarrow \dfrac{\text{sin θ}}{\text{sin θ} \times \text{cos θ}}\\[1em] \Rightarrow \dfrac{1}{\text{cos θ}}\\[1em] \Rightarrow \text{sec θ}.$

∴ Reason (R) is true.

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

Hence, option 4 is the correct option.

Assertion (A): If 0° < A + B ≤ 90°, A > B and cos (A + B) = $\dfrac{1}{2}$ = sin (A - B), then we can say that A = 45° and B = 15°.

Reason (R): sin 60° = cos 60°.

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, cos (A + B) = $\dfrac{1}{2}$

Since 0° < A + B ≤ 90°, the angle whose cosine = $\dfrac{1}{2}$ is 60°.

So, A + B = 60° ....................(1)

sin (A - B) = $\dfrac{1}{2}$

Since A > B, A - B will be positive. The angle whose sine = $\dfrac{1}{2}$ is 30°.

So, A - B = 30° ....................(2)

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

⇒ (A + B) + (A - B) = 60° + 30°

⇒ A + B + A - B = 90°

⇒ 2A = 90°

⇒ A = $\dfrac{90°}{2}$

⇒ A = 45°

Substituting the value of A in equation (1), we get :

⇒ 45° + B = 60°

⇒ B = 60° - 45°

⇒ B = 15°.

∴ Assertion (A) is true.

sin 60° = $\dfrac{\sqrt{3}}{2}$

cos 60° = $\dfrac{1}{2}$

As sin 60° ≠ cos 60°

∴ Reason (R) is false.

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

Hence, option 1 is the correct option.

Assertion (A): If ΔABC is equilateral, then cos A + cos B + cos C = sin A + sin B + sin C.

Reason (R): In isosceles right angled triangle ABC, cos A + cos C = sin A + sin C.

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

According to assertion, if ΔABC is equilateral, then cos A + cos B + cos C = sin A + sin B + sin C.

In an equilateral triangle, all three angles are equal to 60°. So, A = B = C = 60°.

Taking L.H.S.,

⇒ cos A + cos B + cos C

⇒ cos 60° + cos 60° + cos 60°

⇒ $\dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2}$

⇒ $\dfrac{3}{2}$

Taking R.H.S.,

⇒ sin A + sin B + sin C

⇒ sin 60° + sin 60° + sin 60°

⇒ $\dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{3}}{2}$

⇒ $\dfrac{3\sqrt{3}}{2}$

As L.H.S. ≠ R.H.S.

∴ Assertion (A) is false.

According to reason, in isosceles right angled triangle ABC, cos A + cos C = sin A + sin C.

An isosceles right-angled triangle has angles 45°, 45°, and 90°. In triangle ABC, the right angle is at B, so ∠B = 90°. Then the two equal angles are A = 45° and C = 45°.

Taking L.H.S.,

⇒ cos A + cos C

⇒ cos 45° + cos 45°

⇒ $\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}}$

⇒ $\dfrac{2}{\sqrt{2}} = \sqrt{2}$

Taking R.H.S.,

⇒ sin A + sin C

⇒ sin 45° + sin 45°

⇒ $\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}}$

⇒ $\dfrac{2}{\sqrt{2}} = \sqrt{2}$

As L.H.S. = R.H.S.

∴ Reason (R) is true.

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

Hence, option 2 is the correct option.