`sec 2x - tan 2x=`

Options

A. `tan (x-pi/4)`
B. `tan(pi/4 -x)`
C. `tan(x + pi/4)`
D. `tan^(2)(pi/4 +x)`

Answer

To solve the expression \( \sec 2x - \tan 2x \), we can follow these steps:

### Step 1: Write the expression in terms of sine and cosine
We know that:
\[ \sec 2x = \frac{1}{\cos 2x} \quad \text{and} \quad \tan 2x = \frac{\sin 2x}{\cos 2x} \]
Thus, we can rewrite the expression:
\[ \sec 2x - \tan 2x = \frac{1}{\cos 2x} - \frac{\sin 2x}{\cos 2x} \]

### Step 2: Combine the fractions
Since both terms have the same denominator, we can combine them:
\[ \sec 2x - \tan 2x = \frac{1 - \sin 2x}{\cos 2x} \]

### Step 3: Substitute the double angle formulas
Using the double angle formulas, we know:
\[ \sin 2x = 2 \sin x \cos x \quad \text{and} \quad \cos 2x = \cos^2 x - \sin^2 x \]
Substituting these into our expression gives:
\[ \sec 2x - \tan 2x = \frac{1 - 2 \sin x \cos x}{\cos 2x} \]

### Step 4: Rewrite the denominator
We can express \( \cos 2x \) in terms of \( \sin x \) and \( \cos x \):
\[ \cos 2x = \cos^2 x - \sin^2 x = (1 - \sin^2 x) - \sin^2 x = 1 - 2\sin^2 x \]
Thus, we have:
\[ \sec 2x - \tan 2x = \frac{1 - 2 \sin x \cos x}{1 - 2 \sin^2 x} \]

### Step 5: Factor the numerator
The numerator \( 1 - 2 \sin x \cos x \) can be factored using the identity for sine:
\[ 1 - 2 \sin x \cos x = (1 - \sin 2x) \]
So we can rewrite our expression as:
\[ \sec 2x - \tan 2x = \frac{1 - \sin 2x}{1 - 2 \sin^2 x} \]

### Step 6: Simplify the expression
Now we can simplify further if needed, but for the purpose of this problem, we can leave it in this form.

### Final Result
Thus, the final expression for \( \sec 2x - \tan 2x \) is:
\[ \sec 2x - \tan 2x = \frac{1 - \sin 2x}{\cos 2x} \]