Certainly! Let's go through the steps to find \(\sec \theta + \tan \theta\) given that \(\sin \theta = \frac{a}{b}\). Here, we'll consider \(a\) and \(b\) as example values to demonstrate the process.
1. Given:
[tex]\[
\sin \theta = \frac{a}{b}
\][/tex]
2. Determine \(\cos \theta\):
Using the Pythagorean identity:
[tex]\[
\sin^2 \theta + \cos^2 \theta = 1
\][/tex]
Substituting \(\sin \theta\):
[tex]\[
\left(\frac{a}{b}\right)^2 + \cos^2 \theta = 1
\][/tex]
Simplify:
[tex]\[
\frac{a^2}{b^2} + \cos^2 \theta = 1
\][/tex]
Rearrange to solve for \(\cos^2 \theta\):
[tex]\[
\cos^2 \theta = 1 - \frac{a^2}{b^2}
\][/tex]
Simplify further:
[tex]\[
\cos^2 \theta = \frac{b^2 - a^2}{b^2}
\][/tex]
Taking the square root of both sides:
[tex]\[
\cos \theta = \sqrt{\frac{b^2 - a^2}{b^2}} = \frac{\sqrt{b^2 - a^2}}{b}
\][/tex]
3. Calculate \(\sec \theta\):
Recall that \(\sec \theta = \frac{1}{\cos \theta}\):
[tex]\[
\sec \theta = \frac{1}{\frac{\sqrt{b^2 - a^2}}{b}} = \frac{b}{\sqrt{b^2 - a^2}}
\][/tex]
4. Calculate \(\tan \theta\):
Recall that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\):
[tex]\[
\tan \theta = \frac{\frac{a}{b}}{\frac{\sqrt{b^2 - a^2}}{b}} = \frac{a}{\sqrt{b^2 - a^2}}
\][/tex]
5. Add \(\sec \theta\) and \(\tan \theta\):
Combine the expressions for \(\sec \theta\) and \(\tan \theta\):
[tex]\[
\sec \theta + \tan \theta = \frac{b}{\sqrt{b^2 - a^2}} + \frac{a}{\sqrt{b^2 - a^2}}
\][/tex]
Since the denominators are the same, you can combine the fractions:
[tex]\[
\sec \theta + \tan \theta = \frac{b + a}{\sqrt{b^2 - a^2}}
\][/tex]
Therefore, given \(\sin \theta = \frac{a}{b}\), the expression for \(\sec \theta + \tan \theta\) in terms of \(a\) and \(b\) is:
[tex]\[
\sec \theta + \tan \theta = \frac{b + a}{\sqrt{b^2 - a^2}}
\][/tex]
