If `tanalpha=2`, then the value of `(sinalpha)/(sin^(3)alpha+cos^(3)alpha)` is

Options

A. `(2)/(9)`
B. `(sqrt(5))/(9)`
C. `(10)/(9)`
D. `(5sqrt(5))/(9)`

Answer

To solve the problem, we need to find the value of \(\frac{\sin \alpha}{\sin^3 \alpha + \cos^3 \alpha}\) given that \(\tan \alpha = 2\).

### Step-by-Step Solution:

1. **Understanding the given information:**
We know that \(\tan \alpha = 2\). This means that in a right triangle, the opposite side (perpendicular) is 2 and the adjacent side (base) is 1.

2. **Finding the hypotenuse:**
Using the Pythagorean theorem:
\[ \text{Hypotenuse} = \sqrt{(\text{opposite})^2 + (\text{adjacent})^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \]

3. **Calculating \(\sin \alpha\) and \(\cos \alpha\):**
- \(\sin \alpha = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{2}{\sqrt{5}}\)
- \(\cos \alpha = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{5}}\)

4. **Calculating \(\sin^3 \alpha\) and \(\cos^3 \alpha\):**
- \(\sin^3 \alpha = \left(\frac{2}{\sqrt{5}}\right)^3 = \frac{8}{5\sqrt{5}}\)
- \(\cos^3 \alpha = \left(\frac{1}{\sqrt{5}}\right)^3 = \frac{1}{5\sqrt{5}}\)

5. **Finding \(\sin^3 \alpha + \cos^3 \alpha\):**
\[ \sin^3 \alpha + \cos^3 \alpha = \frac{8}{5\sqrt{5}} + \frac{1}{5\sqrt{5}} = \frac{8 + 1}{5\sqrt{5}} = \frac{9}{5\sqrt{5}} \]

6. **Substituting into the expression:**
Now we substitute \(\sin \alpha\) and \(\sin^3 \alpha + \cos^3 \alpha\) into the expression:
\[ \frac{\sin \alpha}{\sin^3 \alpha + \cos^3 \alpha} = \frac{\frac{2}{\sqrt{5}}}{\frac{9}{5\sqrt{5}}} \]

7. **Simplifying the expression:**
\[ = \frac{2}{\sqrt{5}} \times \frac{5\sqrt{5}}{9} = \frac{2 \cdot 5}{9} = \frac{10}{9} \]

### Final Answer:
The value of \(\frac{\sin \alpha}{\sin^3 \alpha + \cos^3 \alpha}\) is \(\frac{10}{9}\).

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