$$ 4.3 .2 \frac { \sin 120 ^ { \circ } \cdot an 300 ^ { \circ } } { \cos ( - 60 ^ { \circ } ) \cdot an 135 } $$

Question

$$ 4.3 .2 \frac { \sin 120 ^ { \circ } \cdot 	an 300 ^ { \circ } } { \cos ( - 60 ^ { \circ } ) \cdot 	an 135 } $$

UpStudy AI · Accepted Answer

We are given the expression

$$
\frac{\sin 120^\circ \cdot \tan 300^\circ}{\cos(-60^\circ) \cdot \tan 135^\circ}.
$$

Let’s evaluate it step by step.

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 $$\tan 300^\circ$$:

Notice that
   $$
   300^\circ = 360^\circ-60^\circ.
   $$
   The tangent function has period $$180^\circ$$ and is odd in the sense that $$\tan(360^\circ-\theta) = -\tan \theta$$. Thus,
   $$
   \tan 300^\circ = -\tan 60^\circ.
   $$
   Since
   $$
   \tan 60^\circ = \sqrt{3},
   $$
   it follows that
   $$
   \tan 300^\circ = -\sqrt{3}.
   $$

3. Evaluate $$\cos(-60^\circ)$$:

The cosine function is even, so
   $$
   \cos(-60^\circ)=\cos 60^\circ.
   $$
   And we know
   $$
   \cos 60^\circ=\frac{1}{2}.
   $$

4. Evaluate $$\tan 135^\circ$$:

We have
   $$
   135^\circ = 180^\circ-45^\circ,
   $$
   so
   $$
   \tan 135^\circ = -\tan 45^\circ.
   $$
   Since
   $$
   \tan 45^\circ=1,
   $$
   it follows that
   $$
   \tan 135^\circ = -1.
   $$

5. Substitute all values into the expression:

The numerator becomes
   $$
   \sin 120^\circ \cdot \tan 300^\circ = \frac{\sqrt{3}}{2} \cdot \left(-\sqrt{3}\right) = -\frac{3}{2}.
   $$
   The denominator becomes
   $$
   \cos(-60^\circ) \cdot \tan 135^\circ = \frac{1}{2} \cdot (-1) = -\frac{1}{2}.
   $$

6. Compute the fraction:

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

Thus, the final answer is

$$
3.
$$
The final answer is 3.

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