Question 4

We will check each statement using trigonometric definitions and specific angle values.

Step 1 — Check statement (i)

Let's look at the first statement. It says $\sin (A + B) = \sin A + \sin B$. We need to see if this is always true. Counterexample method: To prove a statement is false, we only need one specific case where it fails.

Let's pick some angles for $\mathbf{A}$ and $\mathbf{B}$. Let $\mathbf{A = 30^\circ}$ and $\mathbf{B = 60^\circ}$. First, we find $\sin (A + B)$.

$\sin (A + B) = \sin (30^\circ + 60^\circ)$

$= \sin 90^\circ$

$\boxed{1}$

Next, we find $\sin A + \sin B$.

$\sin A + \sin B = \sin 30^\circ + \sin 60^\circ$

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

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

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

We see that $\mathbf{1}$ is not equal to $\mathbf{\frac{1 + \sqrt{3}}{2}}$. So, the statement is false.

Step 2 — Check statement (ii)

Now, let's check the second statement. It says the value of $\sin \theta$ increases as $\theta$ increases. Why sin increases: As angle θ grows from 0° to 90°, the opposite side gets longer relative to the hypotenuse in a right triangle — so sin θ = Opp/Hyp increases from 0 to 1.

We usually consider angles from $\mathbf{0^\circ}$ to $\mathbf{90^\circ}$ in Class 9. Let's look at some values of $\sin \theta$. For $\theta = \mathbf{0^\circ}$:

$\sin 0^\circ$

$\boxed{0}$

For $\theta = \mathbf{30^\circ}$:

$\sin 30^\circ$

$\boxed{\frac{1}{2}}$

For $\theta = \mathbf{45^\circ}$:

$\sin 45^\circ$

$\boxed{\frac{1}{\sqrt{2}}}$

For $\theta = \mathbf{60^\circ}$:

$\sin 60^\circ$

$\boxed{\frac{\sqrt{3}}{2}}$

For $\theta = \mathbf{90^\circ}$:

$\sin 90^\circ$

$\boxed{1}$

We can see that $\mathbf{0 < \frac{1}{2} < \frac{1}{\sqrt{2}} < \frac{\sqrt{3}}{2} < 1}$. The values are increasing. So, for angles from $\mathbf{0^\circ}$ to $\mathbf{90^\circ}$, the statement is true.

Step 3 — Check statement (iii)

Let's check the third statement. It says the value of $\cos \theta$ increases as $\theta$ increases. Why cos decreases: As θ increases, the adjacent side gets shorter relative to the hypotenuse — so cos θ = Adj/Hyp decreases from 1 to 0.

Again, we consider angles from $\mathbf{0^\circ}$ to $\mathbf{90^\circ}$. Let's look at some values of $\cos \theta$. For $\theta = \mathbf{0^\circ}$:

$\cos 0^\circ$

$\boxed{1}$

For $\theta = \mathbf{30^\circ}$:

$\cos 30^\circ$

$\boxed{\frac{\sqrt{3}}{2}}$

For $\theta = \mathbf{45^\circ}$:

$\cos 45^\circ$

$\boxed{\frac{1}{\sqrt{2}}}$

For $\theta = \mathbf{60^\circ}$:

$\cos 60^\circ$

$\boxed{\frac{1}{2}}$

For $\theta = \mathbf{90^\circ}$:

$\cos 90^\circ$

$\boxed{0}$

We can see that $\mathbf{1 > \frac{\sqrt{3}}{2} > \frac{1}{\sqrt{2}} > \frac{1}{2} > 0}$. The values are decreasing. So, for angles from $\mathbf{0^\circ}$ to $\mathbf{90^\circ}$, the statement is false.

Step 4 — Check statement (iv)

Now, let's check the fourth statement. It says $\sin \theta = \cos \theta$ for all values of $\theta$. We need to find if this is always true. Let's pick an angle, for example, $\mathbf{\theta = 30^\circ}$. First, we find $\sin 30^\circ$.

$\sin 30^\circ$

$\boxed{\frac{1}{2}}$

Next, we find $\cos 30^\circ$.

$\cos 30^\circ$

$\boxed{\frac{\sqrt{3}}{2}}$

We see that $\mathbf{\frac{1}{2}}$ is not equal to $\mathbf{\frac{\sqrt{3}}{2}}$. They are only equal for $\mathbf{\theta = 45^\circ}$. So, the statement is false.

Step 5 — Check statement (v)

Finally, let's check the fifth statement. It says $\cot A$ is not defined for $A = 0^\circ$. Undefined ratios: Any trig ratio with 0 in the denominator is undefined. $\cot A = \frac{\cos A}{\sin A}$, so it is undefined whenever $\sin A = 0$, which happens at $A = 0°$.

We know that $\cot A$ is defined as $\frac{\cos A}{\sin A}$. Let's substitute $\mathbf{A = 0^\circ}$ into this definition.

$\cot 0^\circ = \frac{\cos 0^\circ}{\sin 0^\circ}$

$= \frac{1}{0}$

$\boxed{\text{Undefined}}$

Division by zero is not allowed. So, $\cot A$ is indeed not defined for $\mathbf{A = 0^\circ}$. The statement is true.

Answer

(i) False (ii) True (iii) False (iv) False (v) True