`lim_(x to pi/2)(1-sinx)/(cosx)` is equal to :

Options

A. 0
B. -1
C. 1
D. does not exist

Answer

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

### Step 1: Substitute \( x = \frac{\pi}{2} \)
First, we substitute \( x = \frac{\pi}{2} \) into the expression to check if it results in an indeterminate form.

\[ 1 - \sin\left(\frac{\pi}{2}\right) = 1 - 1 = 0 \]
\[ \cos\left(\frac{\pi}{2}\right) = 0 \]

Thus, we have:

\[ \frac{0}{0} \]

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator.

The numerator is \( 1 - \sin x \) and the denominator is \( \cos x \).

- The derivative of the numerator \( 1 - \sin x \) is \( -\cos x \).
- The derivative of the denominator \( \cos x \) is \( -\sin x \).

Thus, we can rewrite the limit as:

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

### Step 3: Simplify the Limit
Now we simplify the limit:

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

### Step 4: Substitute \( x = \frac{\pi}{2} \) Again
Now we substitute \( x = \frac{\pi}{2} \) again:

\[ \cos\left(\frac{\pi}{2}\right) = 0 \]
\[ \sin\left(\frac{\pi}{2}\right) = 1 \]

Thus, we have:

\[ \frac{0}{1} = 0 \]

### Final Answer
Therefore, the limit is:

\[ \lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\cos x} = 0 \]

---