State whether the following are true or false. Justify your answer.

(i) sin (A + B) = sin A + sin B.

(ii) The value of sin θ increases as θ increases.

(iii) The value of cos θ increases as θ increases.

(iv) sin θ = cos θ for all values of θ.

(v) cot A is not defined for A = 0°.

(i) Let A = 30° and B = 60°.

Substituting value of A and B in L.H.S. of the equation sin (A + B) = sin A + sin B, we get :

⇒ sin(A + B)

⇒ sin(30° + 60°)

⇒ sin 90°

⇒ 1.

Substituting value of A and B in R.H.S. of the equation sin (A + B) = sin A + sin B, we get :

⇒ sin A + sin B

⇒ sin 30° + sin 60°

⇒ $\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}$

⇒ $\dfrac{1 + \sqrt{3}}{2}$.

Since, sin (A + B) ≠ sin A + sin B.

Hence, statement sin (A + B) = sin A + sin B is false.

(ii) Let θ be 0°, 30° and 45°.

sin 0° = 0, sin 30° = $\dfrac{1}{2}$ and sin 45° = $\dfrac{1}{\sqrt{2}}$.

We see that,

sin θ increases as θ increases.

Hence, the statement the value of sin θ increases as θ increases is true.

(iii) Let θ be 0°, 30° and 45°.

cos 0° = 1, cos 30° = $\dfrac{\sqrt{3}}{2}$ and cos 45° = $\dfrac{1}{\sqrt{2}}$.

We see that,

cos θ decreases as θ increases.

Hence, the statement the value of cos θ increases as θ increases is false.

(iv) Let θ be 0°, 30° and 45°.

sin 0° = 0, sin 30° = $\dfrac{1}{2}$ and sin 45° = $\dfrac{1}{\sqrt{2}}$,

cos 0° = 1, cos 30° = $\dfrac{\sqrt{3}}{2}$ and sin 45° = $\dfrac{1}{\sqrt{2}}$,

From above we see that,

sin θ = cos θ, only when θ = 45°.

Hence, the statement sin θ = cos θ for all values of θ is false.

(v) We know that,

cot A = $\dfrac{\text{cos A}}{\text{sin A}}$

For A = 0°, sin A = 0.

∴ cot A is not defined.

Hence, the statement cot A is not defined for A = 0° is true.