Certainly! Let's break this down step by step and complete the identity:
We start with the given expression:
[tex]\[
\sin^2 x + \tan^2 x + \cos^2 x
\][/tex]
First, recall two fundamental trigonometric identities:
1. \(\sin^2 x + \cos^2 x = 1\)
2. \(\tan x = \frac{\sin x}{\cos x}\), so \(\tan^2 x = \left(\frac{\sin x}{\cos x}\right)^2 = \frac{\sin^2 x}{\cos^2 x}\)
Using these identities, we can rewrite the given expression:
[tex]\[
\sin^2 x + \tan^2 x + \cos^2 x = \sin^2 x + \frac{\sin^2 x}{\cos^2 x} + \cos^2 x
\][/tex]
We simplify the expression by recognizing that the second term can be written using the identity of secant (\(\sec x = \frac{1}{\cos x}\)):
[tex]\[
\sin^2 x + \frac{\sin^2 x}{\cos^2 x} + \cos^2 x = \sin^2 x + \sin^2 x \cdot \sec^2 x + \cos^2 x
\][/tex]
Next, substitute \(\sec^2 x\) with its identity \(\sec^2 x = 1 + \tan^2 x\):
[tex]\[
\sin^2 x + \sin^2 x (1 + \tan^2 x) + \cos^2 x
\][/tex]
Now we simplify the product:
[tex]\[
\sin^2 x + \sin^2 x + \sin^2 x \tan^2 x + \cos^2 x = \sin^2 x + \sin^2 x \tan^2 x + \cos^2 x + \sin^2 x
\][/tex]
Combine like terms. Notice \(\sin^2 x\) appears twice:
[tex]\[
2 \sin^2 x + \cos^2 x + \sin^2 x \tan^2 x
\][/tex]
Since \(\sin^2 x + \cos^2 x = 1\):
[tex]\[
1 + \sin^2 x \tan^2 x
\][/tex]
We now notice that \(\sec^2 x = 1 + \tan^2 x\):
So we equate:
[tex]\[
1 + \tan^2 x \quad \text{recall earlier simplification matches} 1 + \sin^2 x
\][/tex]
Thus, the correct identity is:
[tex]\[
\sin^2 x + \tan^2 x + \cos^2 x = \sec^2 x
\][/tex]
The correct answer is:
B. [tex]\(\sec^2 x\)[/tex]
