Answer :

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]

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