If `theta in R - {(4n + 1) pi/4 , n in I}` , then range of `(cos 3 theta + sin 3 theta)/(cos theta - sin theta)` is :

Options

A. `[-1, 3]`
B. `[-1, 3)`
C. `[1, 3)`
D. `[-2,2]`

Explanation

To solve the problem, we need to find the range of the function

\[ f(\theta) = \frac{\cos 3\theta + \sin 3\theta}{\cos \theta - \sin \theta} \]

given that \(\theta \in \mathbb{R} \setminus \{(4n + 1)\frac{\pi}{4} : n \in \mathbb{Z}\}\).

### Step 1: Rewrite \(\cos 3\theta + \sin 3\theta\)

Using the angle addition formulas, we can express \(\cos 3\theta\) and \(\sin 3\theta\):

\[ \cos 3\theta = 4\cos^3 \theta - 3\cos \theta \]
\[ \sin 3\theta = 3\sin \theta - 4\sin^3 \theta \]

Thus,

\[ \cos 3\theta + \sin 3\theta = (4\cos^3 \theta - 3\cos \theta) + (3\sin \theta - 4\sin^3 \theta) \]

### Step 2: Simplify the numerator

Combining the terms, we have:

\[ \cos 3\theta + \sin 3\theta = 4\cos^3 \theta - 4\sin^3 \theta + 3\sin \theta - 3\cos \theta \]

Factoring out \(4\) from the first two terms gives:

\[ = 4(\cos^3 \theta - \sin^3 \theta) + 3(\sin \theta - \cos \theta) \]

Using the identity for the difference of cubes, we get:

\[ \cos^3 \theta - \sin^3 \theta = (\cos \theta - \sin \theta)(\cos^2 \theta + \cos \theta \sin \theta + \sin^2 \theta) \]

Since \(\cos^2 \theta + \sin^2 \theta = 1\), we can simplify further:

\[ = (\cos \theta - \sin \theta)(1 + \cos \theta \sin \theta) \]

Thus, the numerator becomes:

\[ \cos 3\theta + \sin 3\theta = 4(\cos \theta - \sin \theta)(1 + \cos \theta \sin \theta) + 3(\sin \theta - \cos \theta) \]

### Step 3: Substitute into the function

Now substituting back into \(f(\theta)\):

\[ f(\theta) = \frac{4(\cos \theta - \sin \theta)(1 + \cos \theta \sin \theta) + 3(\sin \theta - \cos \theta)}{\cos \theta - \sin \theta} \]

### Step 4: Cancel the common terms

Since \(\cos \theta - \sin \theta\) is not zero (as \(\theta\) does not belong to the excluded set), we can cancel:

\[ f(\theta) = 4(1 + \cos \theta \sin \theta) - 3 \]

This simplifies to:

\[ f(\theta) = 1 + 4\cos \theta \sin \theta \]

### Step 5: Use the double angle identity

Using the double angle identity \(2\sin \theta \cos \theta = \sin 2\theta\):

\[ f(\theta) = 1 + 2\sin 2\theta \]

### Step 6: Determine the range of \(f(\theta)\)

The sine function \(\sin 2\theta\) ranges from \(-1\) to \(1\). Therefore, \(2\sin 2\theta\) ranges from \(-2\) to \(2\).

Adding \(1\) to this range gives:

\[ 1 + 2\sin 2\theta \text{ ranges from } 1 - 2 = -1 \text{ to } 1 + 2 = 3 \]

### Step 7: Exclude the point where \(\theta = \frac{\pi}{4}\)

Since \(\theta\) cannot take the values of \((4n + 1)\frac{\pi}{4}\), we need to check if \(f(\theta)\) can equal \(3\):

Setting \(f(\theta) = 3\):

\[ 1 + 2\sin 2\theta = 3 \implies 2\sin 2\theta = 2 \implies \sin 2\theta = 1 \]

This occurs when \(2\theta = \frac{\pi}{2} + 2k\pi\) or \(\theta = \frac{\pi}{4} + k\pi\), which is excluded from the domain.

### Final Range

Thus, the range of \(f(\theta)\) is:

\[ \text{Range of } f(\theta) = (-1, 3) \]