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

1. Problem Description

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

2. Solution Steps

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

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

We also know the trigonometric identity:

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

Substituting $sin \alpha = 5cos \alpha$ into the identity, we get:

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

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

$26cos^2 \alpha = 1$

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

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

Since $sin \alpha = 5cos \alpha$, we have $sin \alpha = \pm \frac{5}{\sqrt{26}}$.

Now, we want to find $sin \alpha \cdot cos \alpha$:

$sin \alpha \cdot cos \alpha = (5cos \alpha) \cdot cos \alpha = 5cos^2 \alpha$

$sin \alpha \cdot cos \alpha = 5 \cdot \frac{1}{26} = \frac{5}{26}$.

Alternatively, if we plug in the value of $sin \alpha$ and $cos \alpha$:

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

3. Final Answer

The value of $sin \alpha \cdot cos \alpha$ is $\frac{5}{26}$.