To solve the given limit using trigonometric identities, the expression is:

$\lim_{{x \to \frac{\pi}{4}}} \left( \frac{\cos(2x)}{\sin(x) - \cos(x)} + \tan(x) + 1 \right)$

Let's break this down step by step using trigonometric identities.

Step 1: Trigonometric identities

1. Double-angle identity for cosine: $\cos(2x) = \cos^2(x) - \sin^2(x)$
2. Simplify the term $\tan(x)$: $\tan(x) = \frac{\sin(x)}{\cos(x)}$

Step 2: Substitute $\cos(2x)$ and simplify

The expression becomes:

$\lim_{{x \to \frac{\pi}{4}}} \left( \frac{\cos^2(x) - \sin^2(x)}{\sin(x) - \cos(x)} + \frac{\sin(x)}{\cos(x)} + 1 \right)$

Step 3: Evaluate at $x = \frac{\pi}{4}$

At $x = \frac{\pi}{4}$, the values of the basic trigonometric functions are:

• $\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
• $\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

Substitute these values into the limit expression:

1. Cosine double angle at $\frac{\pi}{4}$: $\cos(2x) = \cos\left(\frac{\pi}{2}\right) = 0$ So the first term becomes: $\frac{\cos^2\left(\frac{\pi}{4}\right) - \sin^2\left(\frac{\pi}{4}\right)}{\sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right)} = \frac{0}{0}$ This is an indeterminate form, which suggests we need to simplify further.

Step 4: Apply L'Hopital's Rule

Since the first term gives an indeterminate form $\frac{0}{0}$, we apply L'Hopital's Rule by differentiating the numerator and denominator:

• Derivative of the numerator $\cos(2x)$: $\frac{d}{dx}[\cos(2x)] = -2\sin(2x)$
• Derivative of the denominator $\sin(x) - \cos(x)$: $\frac{d}{dx}[\sin(x) - \cos(x)] = \cos(x) + \sin(x)$

So the limit of the first term becomes:

$\lim_{{x \to \frac{\pi}{4}}} \frac{-2\sin(2x)}{\cos(x) + \sin(x)}$

At $x = \frac{\pi}{4}$:

• (\sin\left(\frac{\pi}{2