To find the value of [tex]\(\tan \theta\)[/tex] given [tex]\(\sin \theta = \frac{m}{n}\)[/tex], we need to use some trigonometric identities and relationships. Here is the detailed, step-by-step solution:
### Step 1: Given Information
We are given:
[tex]\[
\sin \theta = \frac{m}{n}
\][/tex]
### Step 2: Pythagorean Identity
We know the Pythagorean identity for sine and cosine:
[tex]\[
\sin^2 \theta + \cos^2 \theta = 1
\][/tex]
### Step 3: Find [tex]\(\cos \theta\)[/tex]
First, square the given value for [tex]\(\sin \theta\)[/tex]:
[tex]\[
\sin^2 \theta = \left(\frac{m}{n}\right)^2 = \frac{m^2}{n^2}
\][/tex]
Next, substitute [tex]\(\sin^2 \theta\)[/tex] into the Pythagorean identity:
[tex]\[
\frac{m^2}{n^2} + \cos^2 \theta = 1
\][/tex]
Solve for [tex]\(\cos^2 \theta\)[/tex]:
[tex]\[
\cos^2 \theta = 1 - \frac{m^2}{n^2}
\][/tex]
Since [tex]\(1\)[/tex] can be written as [tex]\(\frac{n^2}{n^2}\)[/tex], we have:
[tex]\[
\cos^2 \theta = \frac{n^2 - m^2}{n^2}
\][/tex]
Taking the square root of both sides gives us [tex]\(\cos \theta\)[/tex] (considering the principal value for simplicity):
[tex]\[
\cos \theta = \frac{\sqrt{n^2 - m^2}}{n}
\][/tex]
### Step 4: Find [tex]\(\tan \theta\)[/tex]
We know that:
[tex]\[
\tan \theta = \frac{\sin \theta}{\cos \theta}
\][/tex]
Substitute the values we have found for [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
[tex]\[
\tan \theta = \frac{\frac{m}{n}}{\frac{\sqrt{n^2 - m^2}}{n}}
\][/tex]
This simplifies as follows:
[tex]\[
\tan \theta = \frac{m}{\sqrt{n^2 - m^2}}
\][/tex]
### Final Answer
Thus, the value of [tex]\(\tan \theta\)[/tex] is:
[tex]\[
\tan \theta = \frac{m}{\sqrt{n^2 - m^2}}
\][/tex]
