https://r2.erweima.ai/mathbot/image/17268398882253g7o7kj3.jpg

The problem provided in the image is:

Given \(\tan \alpha = 3\), calculate:

\[
S = \tan\left(\alpha + \frac{3\pi}{2}\right) + \cot(3\pi - \alpha)
\]

### Step 1: Simplify each trigonometric expression.

#### Expression 1: \(\tan\left(\alpha + \frac{3\pi}{2}\right)\)

Using the periodicity of the tangent function, we know that:

\[
\tan(x + \pi) = \tan(x)
\]
and

\[
\tan\left(\alpha + \frac{3\pi}{2}\right) = \tan\left(\alpha + \pi + \frac{\pi}{2}\right) = -\cot(\alpha)
\]

So,

\[
\tan\left(\alpha + \frac{3\pi}{2}\right) = -\cot(\alpha)
\]

#### Expression 2: \(\cot(3\pi - \alpha)\)

Using the identity \(\cot(\pi - x) = -\cot(x)\), we get:

\[
\cot(3\pi - \alpha) = -\cot(\alpha)
\]

### Step 2: Substitute and simplify.

Now, substitute the simplified expressions into the equation for \(S\):

\[
S = -\cot(\alpha) + (-\cot(\alpha)) = -2\cot(\alpha)
\]

### Step 3: Use the given \(\tan \alpha = 3\).

We know that:

\[
\cot(\alpha) = \frac{1}{\tan(\alpha)} = \frac{1}{3}
\]

Thus,

\[
S = -2 \times \frac{1}{3} = -\frac{2}{3}
\]

### Final Answer:

\[
S = -\frac{2}{3}
\]

---

Would you like further clarification or details on any part of this solution?

Here are five related questions you might find useful:
1. How can we derive the cotangent function from the tangent function?
2. What are the periodicity properties of trigonometric functions like sine and tangent?
3. Can you explain why \(\tan(x + \pi) = \tan(x)\)?
4. How does the cotangent identity \(\cot(\pi - x) = -\cot(x)\) work geometrically?
5. How do phase shifts in trigonometric functions affect their graphs?

**Tip:** Understanding the symmetry and periodicity of trigonometric functions helps in solving problems that involve phase shifts and transformations.

Given tan(α) = 3, calculate: S = tan(α + 3π/2) + cot(3π - α)

Learn how to solve a trigonometric equation using tangent and cotangent identities. The problem involves simplifying the expression S = tan(α + 3π/2) + cot(3π - α) where tan(α) = 3. Step-by-step explanation suitable for Grades 10-12.

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