Certainly! Let's verify the trigonometric identity [tex]\(\sec^2 \theta = 1 + \tan^2 \theta\)[/tex].
### Step-by-Step Solution:
1. Recall Basic Trigonometric Identities:
- The secant function is defined as [tex]\(\sec \theta = \frac{1}{\cos \theta}\)[/tex].
- The tangent function is defined as [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex].
2. Express [tex]\(\sec^2 \theta\)[/tex] in Terms of [tex]\(\cos \theta\)[/tex]:
Since [tex]\(\sec \theta = \frac{1}{\cos \theta}\)[/tex], we can write:
[tex]\[
\sec^2 \theta = \left( \frac{1}{\cos \theta} \right)^2 = \frac{1}{\cos^2 \theta}
\][/tex]
3. Express [tex]\(\tan^2 \theta\)[/tex] in Terms of [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
Since [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex], we can write:
[tex]\[
\tan^2 \theta = \left( \frac{\sin \theta}{\cos \theta} \right)^2 = \frac{\sin^2 \theta}{\cos^2 \theta}
\][/tex]
4. Combine and Simplify:
Substituting these expressions into the right-hand side of the given identity:
[tex]\[
1 + \tan^2 \theta = 1 + \frac{\sin^2 \theta}{\cos^2 \theta}
\][/tex]
To combine these terms over a common denominator, note that 1 can be written as [tex]\(\frac{\cos^2 \theta}{\cos^2 \theta}\)[/tex]:
[tex]\[
1 + \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{\cos^2 \theta}{\cos^2 \theta} + \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta}
\][/tex]
5. Use the Pythagorean Identity:
Recall the Pythagorean identity:
[tex]\[
\sin^2 \theta + \cos^2 \theta = 1
\][/tex]
Substituting this into the equation, we get:
[tex]\[
\frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta}
\][/tex]
6. Compare Both Sides:
We now have:
[tex]\[
1 + \tan^2 \theta = \frac{1}{\cos^2 \theta}
\][/tex]
And:
[tex]\[
\sec^2 \theta = \frac{1}{\cos^2 \theta}
\][/tex]
Since both sides of the equation equal [tex]\(\frac{1}{\cos^2 \theta}\)[/tex], we have proven that the identity holds true.
### Conclusion:
We have verified the trigonometric identity [tex]\(\sec^2 \theta = 1 + \tan^2 \theta\)[/tex] through a series of logical steps and using fundamental trigonometric identities. Therefore, the identity is indeed valid and true.
