If p = $tan^2x+cot^2x$, then which of the following is correct ?

Question

Accepted Answer

To determine why Option 2 ($p \ge 2$) is correct, let's analyze the expression $ p = \tan^2x + \cot^2x $.

### Explanation:

1. **Expression Analysis:**
   - We know that $\tan x = \frac{\sin x}{\cos x}$ and $\cot x = \frac{\cos x}{\sin x}$.
   - Therefore, $\tan^2 x = \frac{\sin^2 x}{\cos^2 x}$ and $\cot^2 x = \frac{\cos^2 x}{\sin^2 x}$.

2. **Sum of Squares:**
   - The expression becomes:
     $$
     p = \frac{\sin^2 x}{\cos^2 x} + \frac{\cos^2 x}{\sin^2 x}
     $$
   - By using the identity $a^2 + b^2 \geq 2ab$ (which is derived from the AM-GM inequality), we can apply it to $\frac{\sin^2 x}{\cos^2 x}$ and $\frac{\cos^2 x}{\sin^2 x}$.

3. **Applying AM-GM Inequality:**
   - Let $a = \frac{\sin^2 x}{\cos^2 x}$ and $b = \frac{\cos^2 x}{\sin^2 x}$.
   - Then, $a \cdot b = \left(\frac{\sin^2 x}{\cos^2 x}\right) \cdot \left(\frac{\cos^2 x}{\sin^2 x}\right) = 1$.
   - According to the AM-GM inequality:
     $$
     a + b \geq 2\sqrt{ab} = 2\sqrt{1} = 2
     $$
   - Therefore, $p = a + b \geq 2$.

4. **Conclusion:**
   - This shows that the minimum value of $p$ is 2, hence $p \geq 2$.

### Why Other Options are Incorrect:

1. **Option 1 ($p \le 2$):**
   - This option suggests that $p$ can be less than or equal to 2. However, as shown above, the minimum value of $p$ is 2, and it cannot be less than 2. Therefore, this option is incorrect.

2. **Option 3 (p < 2):**
   - This option implies that $p$ can be less than 2, which contradicts the derived inequality $p \geq 2$. Hence, this option is incorrect.

3. **Option 4 (p > 2):**
   - While $p$ can indeed be greater than 2, this option does not account for the possibility of $p$ being exactly 2. Therefore, it is not as comprehensive as Option 2, which correctly states $p \geq 2$.

In conclusion, the correct answer is Option 2 ($p \ge 2$) because it accurately reflects the mathematical analysis of the expression $p = \tan^2x + \cot^2x$.