Consider the following :
1. `tan^(2) theta - sin^(2) theta = tan^(2) theta sin ^(2) theta `
2. `(1 + cot^(2) theta) (1 - cos theta) (1 + cos theta) = 1 `
Which of the statements given below is correct

Options

A. 1 only is the identiy
B. 2 only is the identity
C. Both 1 and 2 are identity
D. Neither 1 nor 2 is the identity

Answer

To determine the correctness of the given statements, we will analyze each statement step by step.

### Statement 1:
**tan²θ - sin²θ = tan²θ * sin²θ**

1. **Start with the left-hand side (LHS)**:
\[ \text{LHS} = \tan^2 \theta - \sin^2 \theta \]

2. **Substitute \(\tan^2 \theta\)**:
\[ \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \]
So,
\[ \text{LHS} = \frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta \]

3. **Take LCM (Least Common Multiple)**:
The LCM of \(\cos^2 \theta\) and 1 is \(\cos^2 \theta\):
\[ \text{LHS} = \frac{\sin^2 \theta - \sin^2 \theta \cos^2 \theta}{\cos^2 \theta} \]

4. **Factor out \(\sin^2 \theta\)**:
\[ \text{LHS} = \frac{\sin^2 \theta (1 - \cos^2 \theta)}{\cos^2 \theta} \]

5. **Use the Pythagorean identity**:
\[ 1 - \cos^2 \theta = \sin^2 \theta \]
Thus,
\[ \text{LHS} = \frac{\sin^2 \theta \cdot \sin^2 \theta}{\cos^2 \theta} = \frac{\sin^4 \theta}{\cos^2 \theta} \]

6. **Now consider the right-hand side (RHS)**:
\[ \text{RHS} = \tan^2 \theta \cdot \sin^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \cdot \sin^2 \theta = \frac{\sin^4 \theta}{\cos^2 \theta} \]

7. **Compare LHS and RHS**:
\[ \text{LHS} = \text{RHS} \]
Therefore, Statement 1 is **true**.

### Statement 2:
**(1 + cot²θ)(1 - cosθ)(1 + cosθ) = 1**

1. **Start with the left-hand side (LHS)**:
\[ \text{LHS} = (1 + \cot^2 \theta)(1 - \cos \theta)(1 + \cos \theta) \]

2. **Use the identity for cot²θ**:
\[ 1 + \cot^2 \theta = \frac{1}{\sin^2 \theta} \]
So,
\[ \text{LHS} = \frac{1}{\sin^2 \theta} (1 - \cos \theta)(1 + \cos \theta) \]

3. **Simplify the product (1 - cosθ)(1 + cosθ)**:
\[ (1 - \cos \theta)(1 + \cos \theta) = 1 - \cos^2 \theta = \sin^2 \theta \]

4. **Substituting back into LHS**:
\[ \text{LHS} = \frac{1}{\sin^2 \theta} \cdot \sin^2 \theta = 1 \]

5. **Conclusion**:
Therefore, Statement 2 is also **true**.

### Final Conclusion:
Both statements are correct.