Let's solve the problem step by step.

Problem:

Given $\tan(90^\circ - \theta) = \frac{1}{2}$, find the value of: $\frac{\sin(90^\circ - \theta)}{1 - \tan\theta} + 2\cos(90^\circ - \theta).$

Step 1: Trigonometric Identities

1. Identity for $\tan(90^\circ - \theta)$: $\tan(90^\circ - \theta) = \cot(\theta).$ So, $\cot(\theta) = \frac{1}{2}$.

2. Relationship between $\cot\theta$ and $\tan\theta$: $\tan(\theta) = \frac{1}{\cot(\theta)} = 2.$

3. Sine and Cosine for $90^\circ - \theta$:

   • $\sin(90^\circ - \theta) = \cos(\theta)$,
   • $\cos(90^\circ - \theta) = \sin(\theta)$.

Step 2: Expressing $\sin(\theta)$ and $\cos(\theta)$

From the Pythagorean identity: $1 + \cot^2(\theta) = \csc^2(\theta).$ Using $\cot(\theta) = \frac{1}{2}$: $1 + \left(\frac{1}{2}\right)^2 = \csc^2(\theta).$ $\csc^2(\theta) = 1 + \frac{1}{4} = \frac{5}{4}.$ $\csc(\theta) = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}.$ Since $\csc(\theta) = \frac{1}{\sin(\theta)}$, we have: $\sin(\theta) = \frac{2}{\sqrt{5}}.$ Using $\cos^2(\theta) = 1 - \sin^2(\theta)$: $\cos^2(\theta) = 1 - \left(\frac{2}{\sqrt{5}}\right)^2 = 1 - \frac{4}{5} = \frac{1}{5}.$ $\cos(\theta) = \frac{1}{\sqrt{5}}.$

Step 3: Evaluate the Expression

Substitute $\sin(90^\circ - \theta) = \cos(\theta)$ and $\cos(90^\circ - \theta) = \sin(\theta)$ into the given expression: $\frac{\sin(90^\circ - \theta)}{1 - \tan(\theta)} + 2\cos(90^\circ - \theta).$ This becomes: $\frac{\cos(\theta)}{1 - \tan(\theta)} + 2\sin(\theta).$ Substitute $\tan(\theta) = 2$, $\cos(\theta) = \frac{1}{\sqrt{5}}$, and $\sin(\theta) = \frac{2}{\sqrt{5}}$: $\frac{\frac{1}{\sqrt{5}}}{1 - 2} + 2\cdot\frac{2}{\sqrt{5}}.$ Simplify: $\frac{\frac{1}{\sqrt{5}}}{-1} + \frac{4}{\sqrt{5}}.$ $-\frac{1}{\sqrt{5}} + \frac{4}{\sqrt{5}} = \frac{-1 + 4}{\sqrt{5}} = \frac{3}{\sqrt{5}}.$

Final Answer:

$\boxed{\frac{3}{\sqrt{5}} \text{ or } \frac{3\sqrt{5}}{5} \text{ (rationalized)}.}$

Would you like further details or assistance with another problem? Here are 5 questions to expand your understanding:

1. How does $\tan(90^\circ - \theta) = \cot(\theta)$ affect calculations?
2. Why is rationalizing the denominator important in final answers?
3. What are the applications of trigonometric identities like these in physics or engineering?
4. How would the solution differ if $\tan(90^\circ - \theta)$ had a negative value?
5. Can you generalize this method to solve for other trigonometric expressions?

Tip: Memorize the complementary angle identities ($\sin(90^\circ - \theta) = \cos(\theta)$, etc.) to quickly transform such problems!