If p = cos x - sin x, q = `(1 - sin ^(3) x)/( 1 - sin x) , r = (1 + cos ^(3) x)/( 1 + cos x)` what is the value of p + q + r ?

Options

A. 0
B. 1
C. 2
D. 3

Answer

To find the value of \( p + q + r \) where \( p = \cos x - \sin x \), \( q = \frac{1 - \sin^3 x}{1 - \sin x} \), and \( r = \frac{1 + \cos^3 x}{1 + \cos x} \), we will simplify each term step by step.

### Step 1: Simplifying \( q \)

Given:
\[ q = \frac{1 - \sin^3 x}{1 - \sin x} \]

We can use the identity for the difference of cubes:
\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]
Here, let \( a = 1 \) and \( b = \sin x \). Therefore:
\[ 1 - \sin^3 x = (1 - \sin x)(1 + \sin^2 x + \sin x) \]

Substituting this back into \( q \):
\[ q = \frac{(1 - \sin x)(1 + \sin^2 x + \sin x)}{1 - \sin x} \]

The \( 1 - \sin x \) terms cancel out (as long as \( \sin x \neq 1 \)):
\[ q = 1 + \sin^2 x + \sin x \]

### Step 2: Simplifying \( r \)

Now, we simplify \( r \):
\[ r = \frac{1 + \cos^3 x}{1 + \cos x} \]

Using the identity for the sum of cubes:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Let \( a = 1 \) and \( b = \cos x \). Therefore:
\[ 1 + \cos^3 x = (1 + \cos x)(1 - \cos x + \cos^2 x) \]

Substituting this back into \( r \):
\[ r = \frac{(1 + \cos x)(1 - \cos x + \cos^2 x)}{1 + \cos x} \]

The \( 1 + \cos x \) terms cancel out (as long as \( \cos x \neq -1 \)):
\[ r = 1 - \cos x + \cos^2 x \]

### Step 3: Adding \( p \), \( q \), and \( r \)

Now we have:
- \( p = \cos x - \sin x \)
- \( q = 1 + \sin^2 x + \sin x \)
- \( r = 1 - \cos x + \cos^2 x \)

Let's add them together:
\[ p + q + r = (\cos x - \sin x) + (1 + \sin^2 x + \sin x) + (1 - \cos x + \cos^2 x) \]

Combining like terms:
1. The \( \cos x \) terms: \( \cos x - \cos x = 0 \)
2. The \( \sin x \) terms: \( -\sin x + \sin x = 0 \)
3. The constant terms: \( 1 + 1 = 2 \)
4. The \( \sin^2 x \) and \( \cos^2 x \) terms: \( \sin^2 x + \cos^2 x = 1 \)

So we have:
\[ p + q + r = 2 + 1 = 3 \]

### Final Answer
The value of \( p + q + r \) is \( 3 \).

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