4.3 Simplify the following fully and without the use of a calculator: $$ 4.3 .1 \frac{\cos \left( heta-180^{\circ} ight) an \left(540^{\circ}- heta ight)}{\sin \left(90^{\circ}+ heta ight)} $$ $$ 4.3 .2 \frac{\sin 120^{\circ} \cdot an 300^{\circ}}{\cos \left(-60^{\circ} ight) \cdot an 135^{\circ}} $$

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4.3 Simplify the following fully and without the use of a calculator: $$ 4.3 .1 \frac{\cos \left(	heta-180^{\circ}ight) 	an \left(540^{\circ}-	hetaight)}{\sin \left(90^{\circ}+	hetaight)} $$ $$ 4.3 .2 \frac{\sin 120^{\circ} \cdot 	an 300^{\circ}}{\cos \left(-60^{\circ}ight) \cdot 	an 135^{\circ}} $$

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$$
\textbf{4.3.1:} \quad \frac{\cos(\theta-180^\circ)\,\tan(540^\circ-\theta)}{\sin(90^\circ+\theta)}
$$

1. Simplify $$\cos(\theta-180^\circ)$$:

$$
\cos(\theta-180^\circ) = -\cos\theta
$$

2. Simplify $$\tan(540^\circ-\theta)$$:  
Since tangent has a period of $$180^\circ$$, subtracting $$360^\circ$$ gives

$$
\tan(540^\circ-\theta) = \tan(180^\circ-\theta)
$$

and using the identity

$$
\tan(180^\circ-\theta) = -\tan\theta,
$$

we get

$$
\tan(540^\circ-\theta) = -\tan\theta.
$$

3. Simplify $$\sin(90^\circ+\theta)$$:

$$
\sin(90^\circ+\theta) = \cos\theta.
$$

4. Substitute the simplified expressions into the original expression:

$$
\frac{(-\cos\theta)(-\tan\theta)}{\cos\theta} = \frac{\cos\theta\,\tan\theta}{\cos\theta} = \tan\theta.
$$

Thus, the simplified result is

$$
\boxed{\tan\theta}.
$$

---

$$
\textbf{4.3.2:} \quad \frac{\sin 120^\circ\,\tan 300^\circ}{\cos(-60^\circ)\,\tan 135^\circ}
$$

1. Evaluate $$\sin 120^\circ$$:  
Since

$$
\sin 120^\circ = \sin(180^\circ-60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2},
$$

we have

$$
\sin 120^\circ = \frac{\sqrt{3}}{2}.
$$

2. Evaluate $$\tan 300^\circ$$:  
Express $$300^\circ$$ as $$360^\circ-60^\circ$$. Then

$$
\tan 300^\circ = -\tan 60^\circ = -\sqrt{3}.
$$

3. Evaluate $$\cos(-60^\circ)$$:  
Since cosine is even,

$$
\cos(-60^\circ) = \cos 60^\circ = \frac{1}{2}.
$$

4. Evaluate $$\tan 135^\circ$$:  
Express $$135^\circ$$ as $$180^\circ-45^\circ$$. Then

$$
\tan 135^\circ = -\tan 45^\circ = -1.
$$

5. Substitute these values into the expression:

$$
\frac{\displaystyle \frac{\sqrt{3}}{2} \times (-\sqrt{3})}{\displaystyle \frac{1}{2} \times (-1)}
= \frac{-\frac{3}{2}}{-\frac{1}{2}}.
$$

6. Simplify the fraction:

$$
\frac{-\frac{3}{2}}{-\frac{1}{2}} = \frac{3}{2} \div \frac{1}{2} = 3.
$$

Thus, the simplified result is

$$
\boxed{3}.
$$
$$ \textbf{4.3.1:} \quad \tan\theta $$ $$ \textbf{4.3.2:} \quad 3 $$

4.3 Simplify the following fully and without the use of a calculator:

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