`lim _(xrarr(pi)/(2))(1-sinx)/cos^(2)x`=_____

Answer

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

### Step 1: Write the limit expression
We start with the limit expression:
\[ \lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\cos^2 x} \]

### Step 2: Use the identity for \(\cos^2 x\)
We know the trigonometric identity:
\[ \cos^2 x = 1 - \sin^2 x \]
We can substitute this into our limit expression:
\[ \lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{1 - \sin^2 x} \]

### Step 3: Factor the denominator
The expression \(1 - \sin^2 x\) can be factored using the difference of squares:
\[ 1 - \sin^2 x = (1 - \sin x)(1 + \sin x) \]
Thus, we can rewrite the limit as:
\[ \lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{(1 - \sin x)(1 + \sin x)} \]

### Step 4: Cancel common factors
Since \(1 - \sin x\) appears in both the numerator and the denominator, we can cancel it (as long as \(x \neq \frac{\pi}{2}\)):
\[ \lim_{x \to \frac{\pi}{2}} \frac{1}{1 + \sin x} \]

### Step 5: Substitute the limit
Now we substitute \(x = \frac{\pi}{2}\) into the simplified expression:
\[ \lim_{x \to \frac{\pi}{2}} \frac{1}{1 + \sin\left(\frac{\pi}{2}\right)} = \frac{1}{1 + 1} = \frac{1}{2} \]

### Final Answer
Thus, the limit is:
\[ \frac{1}{2} \]