Evaluate the following :
`lim_(x to pi)(1+cosx)/(tan^(2)x)`

Answer

To evaluate the limit \( \lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x} \), we will follow these steps:

### Step 1: Substitute \( x = \pi \)
First, we substitute \( x = \pi \) into the expression:

\[ 1 + \cos(\pi) = 1 - 1 = 0 \]
\[ \tan^2(\pi) = 0^2 = 0 \]

This gives us the indeterminate form \( \frac{0}{0} \).

**Hint:** When you encounter \( \frac{0}{0} \) form, you can apply L'Hôpital's rule.

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can differentiate the numerator and the denominator.

**Numerator:**
\[ \frac{d}{dx}(1 + \cos x) = -\sin x \]

**Denominator:**
\[ \frac{d}{dx}(\tan^2 x) = 2\tan x \cdot \sec^2 x \]

Now we rewrite the limit using these derivatives:

\[ \lim_{x \to \pi} \frac{-\sin x}{2\tan x \sec^2 x} \]

**Hint:** Remember to simplify the expression after differentiation.

### Step 3: Rewrite \( \tan x \) and \( \sec^2 x \)
Recall that \( \tan x = \frac{\sin x}{\cos x} \) and \( \sec^2 x = \frac{1}{\cos^2 x} \). Thus, we can rewrite the limit:

\[ \lim_{x \to \pi} \frac{-\sin x}{2 \cdot \frac{\sin x}{\cos x} \cdot \frac{1}{\cos^2 x}} = \lim_{x \to \pi} \frac{-\sin x \cdot \cos^2 x}{2 \sin x} \]

**Hint:** Cancel out common terms in the numerator and denominator.

### Step 4: Cancel \( \sin x \)
Assuming \( \sin x \neq 0 \) around \( x = \pi \):

\[ \lim_{x \to \pi} \frac{-\cos^2 x}{2} \]

### Step 5: Substitute \( x = \pi \) again
Now substitute \( x = \pi \):

\[ \cos(\pi) = -1 \quad \Rightarrow \quad \cos^2(\pi) = (-1)^2 = 1 \]

Thus, the limit becomes:

\[ \frac{-1}{2} = -\frac{1}{2} \]

### Final Answer
Therefore, the evaluated limit is:

\[ \lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x} = -\frac{1}{2} \]