Statement-1: `f(theta) = sin^(2)theta+ cos^(2)theta`
then `f(theta)=1` for every real values of `theta`.
Statement:2: `g(theta) = sec^(2)theta - tan^(2)theta`.
then `g(theta) =1` for every real value of `theta`.
Statement-3: `f(theta) = g(theta)` for every real value of `theta`.

Options

A. TFF
B. TTT
C. TFT
D. TTF

Answer

To solve the given statements, we will analyze each statement step by step.

### Step 1: Analyze Statement 1
The first statement is:
\[ f(\theta) = \sin^2 \theta + \cos^2 \theta \]

Using the Pythagorean identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]

Thus, we conclude that:
\[ f(\theta) = 1 \text{ for every real value of } \theta. \]

### Step 2: Analyze Statement 2
The second statement is:
\[ g(\theta) = \sec^2 \theta - \tan^2 \theta \]

Using the identity:
\[ \sec^2 \theta = 1 + \tan^2 \theta \]

We can rewrite \( g(\theta) \):
\[ g(\theta) = (1 + \tan^2 \theta) - \tan^2 \theta = 1 \]

Thus, we conclude that:
\[ g(\theta) = 1 \text{ for every real value of } \theta. \]

### Step 3: Analyze Statement 3
The third statement is:
\[ f(\theta) = g(\theta) \]

From our analysis of Statements 1 and 2, we found:
\[ f(\theta) = 1 \]
\[ g(\theta) = 1 \]

Therefore, we can conclude:
\[ f(\theta) = g(\theta) \text{ for every real value of } \theta. \]

### Conclusion
All three statements are true:
1. \( f(\theta) = 1 \) for every real value of \( \theta \).
2. \( g(\theta) = 1 \) for every real value of \( \theta \).
3. \( f(\theta) = g(\theta) \) for every real value of \( \theta \).

### Final Answer
All three statements are true.

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