$$ \min F 5^{\circ}=\frac{\sin 75^{\circ}}{\cos 75^{\circ}}=\frac{\sqrt{3}+1}{\sqrt{3}-1} $$ P. $$ 2 \cdot+ ightarrow an 8^{\circ}=\frac{\cos 7^{\circ}-\sin 7^{\circ}}{\cos 7^{\circ}+\sin 7^{\circ}} $$

Question

$$ \min F 5^{\circ}=\frac{\sin 75^{\circ}}{\cos 75^{\circ}}=\frac{\sqrt{3}+1}{\sqrt{3}-1} $$ P. $$ 2 \cdot+ightarrow 	an 8^{\circ}=\frac{\cos 7^{\circ}-\sin 7^{\circ}}{\cos 7^{\circ}+\sin 7^{\circ}} $$

UpStudy AI · Accepted Answer

Let's break down the problem step by step.

### Step 1: Analyze the first equation

The first part of the problem states:

$$
F = \frac{\sin 75^{\circ}}{\cos 75^{\circ}} = \tan 75^{\circ}
$$

We know that:

$$
\tan 75^{\circ} = \tan(45^{\circ} + 30^{\circ}) = \frac{\tan 45^{\circ} + \tan 30^{\circ}}{1 - \tan 45^{\circ} \tan 30^{\circ}}
$$

Using the values:

- $$\tan 45^{\circ} = 1$$
- $$\tan 30^{\circ} = \frac{1}{\sqrt{3}}$$

We can substitute these values into the formula:

$$
\tan 75^{\circ} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}}
$$

### Step 2: Simplify the expression

Now, let's simplify the expression:

1. The numerator becomes:
   $$
   1 + \frac{1}{\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3}}
   $$

2. The denominator becomes:
   $$
   1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}}
   $$

Putting it all together:

$$
\tan 75^{\circ} = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}
$$

This confirms the first part of the problem.

### Step 3: Analyze the second equation

The second part of the problem states:

$$
\tan 8^{\circ} = \frac{\cos 7^{\circ} - \sin 7^{\circ}}{\cos 7^{\circ} + \sin 7^{\circ}}
$$

We can use the tangent subtraction formula:

$$
\tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b}
$$

Let $$a = 45^{\circ}$$ and $$b = 37^{\circ}$$ (since $$45^{\circ} - 37^{\circ} = 8^{\circ}$$).

Using the values:

- $$\tan 45^{\circ} = 1$$
- $$\tan 37^{\circ} = \tan(45^{\circ} - 8^{\circ})$$

We can express $$\tan 37^{\circ}$$ in terms of sine and cosine:

$$
\tan 37^{\circ} = \frac{\sin 37^{\circ}}{\cos 37^{\circ}}
$$

### Step 4: Solve for $$\tan 8^{\circ}$$

To find $$\tan 8^{\circ}$$, we can use the identity:

$$
\tan 8^{\circ} = \frac{\tan 45^{\circ} - \tan 37^{\circ}}{1 + \tan 45^{\circ} \tan 37^{\circ}} = \frac{1 - \tan 37^{\circ}}{1 + \tan 37^{\circ}}
$$

Now, we can calculate $$\tan 8^{\circ}$$ using the known values of $$\tan 37^{\circ}$$.

### Conclusion

We have confirmed the first part of the problem and set up the second part. If you would like to proceed with numerical calculations or further simplifications, please let me know!
$$ \tan 75^{\circ} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} $$ $$ \tan 8^{\circ} = \frac{\cos 7^{\circ} - \sin 7^{\circ}}{\cos 7^{\circ} + \sin 7^{\circ}} $$

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