State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.[[1]]
(ii) The value of sin θ increases as θ increases.[[2]]
(iii) The value of cos θ increases as θ increases.[[3]]
(iv) sin θ = cos θ for all values of θ.[[4]]
(v) cot A is not defined for A = 0°.[[5]]

(i) To prove suppose, A = 30° and B = 60°, then

Substitute the values into the given formula, we get

sin (A + B) = sin (30° + 60°) = sin 90° = 1 and,

sin A + sin B = sin 30° + sin 60°

$= \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1+\sqrt{3}}{2}$

Since the obtained values are not equal, the solution is false.

(ii) Writing down all the values of sin as:

$\sin 0° = 0$

$\sin 30° = \frac{1}{2}$

$\sin 45° = \frac{1}{\sqrt{2}}$

$\sin 60° = \frac{\sqrt{3}}{2}$

$\sin 90° = 1$

Since the obtained value of sin θ increases as θ increases, the given statement is true.

(iii) Writing down all the values of cos as:

$\cos 0° = 1$

$\cos 30° = \frac{\sqrt{3}}{2}$

$\cos 45° = \frac{1}{\sqrt{2}}$

$\cos 60° = \frac{1}{2}$

$\cos 90° = 0$

Since the obtained value of cos θ decreases as θ increases, the given statement is false.

(iv) $\sin \theta =\cos \theta =\frac{1}{\sqrt{2}}$

when $\theta =45°$.

So, given statement is true for only $\theta =45°$ and not for all values.

So, the given statement is false.

(v) Since cot function is the reciprocal of the tan function, it can be written as:

$cot A =\frac{1}{\tan A}=\frac{\cos A}{\sin A}$

Now substitute A = 0°

$cot 0° =\frac{\cos 0°}{\sin 0° }=\frac{1}{0}= undefined.$

Thus the given statement is true.