Given $\tan \alpha = 5$, find the value of $\sin \alpha \cdot \cos \alpha$.

1. Problem Description

Given $\tan \alpha = 5$, find the value of $\sin \alpha \cdot \cos \alpha$.

2. Solution Steps

We are given $\tan \alpha = 5$. We want to find the value of $\sin \alpha \cdot \cos \alpha$.

We know that $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$. Therefore, $\frac{\sin \alpha}{\cos \alpha} = 5$, which means $\sin \alpha = 5 \cos \alpha$.

We also know the Pythagorean identity:

$\sin^2 \alpha + \cos^2 \alpha = 1$

Substitute $\sin \alpha = 5 \cos \alpha$ into the identity:

$(5 \cos \alpha)^2 + \cos^2 \alpha = 1$

$25 \cos^2 \alpha + \cos^2 \alpha = 1$

$26 \cos^2 \alpha = 1$

$\cos^2 \alpha = \frac{1}{26}$

So, $\cos \alpha = \pm \frac{1}{\sqrt{26}}$.

Then $\sin \alpha = 5 \cos \alpha = \pm \frac{5}{\sqrt{26}}$.

Finally, $\sin \alpha \cdot \cos \alpha = (\pm \frac{5}{\sqrt{26}}) (\pm \frac{1}{\sqrt{26}}) = \frac{5}{26}$.

Alternatively, we can express $\sin \alpha \cdot \cos \alpha$ in terms of $\tan \alpha$:

$\sin \alpha \cdot \cos \alpha = \frac{\sin \alpha}{\cos \alpha} \cdot \cos^2 \alpha = \tan \alpha \cdot \cos^2 \alpha$

We also have

$1 + \tan^2 \alpha = \frac{1}{\cos^2 \alpha}$

Thus,

$\cos^2 \alpha = \frac{1}{1 + \tan^2 \alpha}$

Then $\sin \alpha \cdot \cos \alpha = \tan \alpha \cdot \cos^2 \alpha = \tan \alpha \cdot \frac{1}{1 + \tan^2 \alpha} = \frac{\tan \alpha}{1 + \tan^2 \alpha}$

Since $\tan \alpha = 5$, $\sin \alpha \cdot \cos \alpha = \frac{5}{1 + 5^2} = \frac{5}{1 + 25} = \frac{5}{26}$.

3. Final Answer

$\frac{5}{26}$