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The expression to simplify is:

\[
\cos\left(\cos^{-1}(u) - \tan^{-1}(v)\right)
\]

### Step-by-step solution:

1. **Recall the inverse cosine and inverse tangent functions:**

   - \(\cos^{-1}(x)\) is the angle whose cosine is \(x\). So, \(\cos^{-1}(u)\) gives an angle \(\theta_1\) such that \(\cos(\theta_1) = u\).
   - \(\tan^{-1}(x)\) is the angle whose tangent is \(x\). So, \(\tan^{-1}(v)\) gives an angle \(\theta_2\) such that \(\tan(\theta_2) = v\).

2. **Substitute trigonometric identities:**

   - Let \(\theta_1 = \cos^{-1}(u)\), so \(\cos(\theta_1) = u\).
   - Let \(\theta_2 = \tan^{-1}(v)\), so \(\tan(\theta_2) = v\).

   Thus, the expression becomes:
   \[
   \cos(\theta_1 - \theta_2)
   \]
   We can now use the cosine difference identity:
   \[
   \cos(\theta_1 - \theta_2) = \cos(\theta_1)\cos(\theta_2) + \sin(\theta_1)\sin(\theta_2)
   \]

3. **Substitute the known values:**

   - From \(\theta_1 = \cos^{-1}(u)\), we know that \(\cos(\theta_1) = u\).
   - From \(\tan(\theta_2) = v\), we can calculate \(\cos(\theta_2)\) and \(\sin(\theta_2)\) using the identity \(\tan(\theta_2) = \frac{\sin(\theta_2)}{\cos(\theta_2)}\). Since \(\tan(\theta_2) = v\), we have:
     \[
     \sin(\theta_2) = \frac{v}{\sqrt{1 + v^2}} \quad \text{and} \quad \cos(\theta_2) = \frac{1}{\sqrt{1 + v^2}}
     \]

   Next, we need \(\sin(\theta_1)\). Since \(\cos(\theta_1) = u\), we use the identity \(\sin^2(\theta_1) + \cos^2(\theta_1) = 1\), so:
   \[
   \sin(\theta_1) = \sqrt{1 - u^2}
   \]

4. **Substitute into the cosine identity:**

   Now substitute all the values:
   \[
   \cos(\theta_1 - \theta_2) = u \cdot \frac{1}{\sqrt{1 + v^2}} + \sqrt{1 - u^2} \cdot \frac{v}{\sqrt{1 + v^2}}
   \]

5. **Simplify the expression:**

   Combine the terms:
   \[
   \cos(\theta_1 - \theta_2) = \frac{u + v\sqrt{1 - u^2}}{\sqrt{1 + v^2}}
   \]

Thus, the final algebraic expression is:

\[
\boxed{\frac{u + v\sqrt{1 - u^2}}{\sqrt{1 + v^2}}}
\]

Would you like any further clarification or details on specific steps? 

Here are 5 related questions for deeper understanding:
1. How does the cosine difference identity work for other trigonometric functions?
2. What is the geometric interpretation of \(\cos^{-1}(x)\) and \(\tan^{-1}(x)\)?
3. Can we derive a similar expression if the problem involved \(\sin^{-1}(x)\)?
4. How do inverse trigonometric functions relate to triangle angles?
5. What are other trigonometric identities useful in simplifying such expressions?

**Tip:** When dealing with inverse trigonometric expressions, always consider rewriting them in terms of standard angles or using trigonometric identities for simplification.

Write the trigonometric expression as an algebraic expression in u and v. Assume that the variables u and v represent positive real numbers. cos(cos⁻¹u - tan⁻¹v)

Learn how to simplify the trigonometric expression cos(cos⁻¹u - tan⁻¹v) as an algebraic expression involving the variables u and v. This problem involves trigonometric identities and inverse functions, suitable for Grades 10-12.

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