Points of Non-Differentiability of $f(x) = |\cos x| + 3$

Step 1: $\cos x$ is differentiable everywhere, but $|\cos x|$ is not differentiable where $\cos x = 0$.

Step 2: In the interval $[-\pi, \pi]$, we have:

$$\cos x = 0 \Rightarrow x = -\frac{\pi}{2},\ \frac{\pi}{2}$$

So $f(x) = |\cos x| + 3$ is not differentiable at these two points due to sharp turns.

✅ Final Answer:   $\boxed{2 \text{ points}}$

We are given:

$$\text{Evaluate } \tan\left(\frac{\pi}{4} + \theta\right) \cdot \tan\left(\frac{3\pi}{4} + \theta\right)$$

✳ Step 1: Use identity

$$\tan\left(A + B\right) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$ But we don’t need expansion — use known angle values:

$$\tan\left(\frac{\pi}{4} + \theta\right) = \frac{1 + \tan\theta}{1 - \tan\theta}$$

$$\tan\left(\frac{3\pi}{4} + \theta\right) = \frac{-1 + \tan\theta}{1 + \tan\theta}$$

✳ Step 2: Multiply

$$\left(\frac{1 + \tan\theta}{1 - \tan\theta}\right) \cdot \left(\frac{-1 + \tan\theta}{1 + \tan\theta}\right)$$

Simplify:

$$= \frac{(1 + \tan\theta)(-1 + \tan\theta)}{(1 - \tan\theta)(1 + \tan\theta)} = \frac{(\tan^2\theta - 1)}{1 - \tan^2\theta} = \boxed{-1}$$

✅ Final Answer:

$$\boxed{-1}$$

Given:

$$\sin x = \sin y \quad \text{and} \quad \cos x = \cos y$$

✳ Step 1: Use the identity for sine

$$\sin x = \sin y \Rightarrow x = y + 2n\pi \quad \text{or} \quad x = \pi - y + 2n\pi$$

✳ Step 2: Use the identity for cosine

$$\cos x = \cos y \Rightarrow x = y + 2m\pi \quad \text{or} \quad x = -y + 2m\pi$$

? Combine both conditions

For both $\sin x = \sin y$ and $\cos x = \cos y$ to be true, the only consistent solution is:

$$x = y + 2n\pi \Rightarrow x - y = 2n\pi$$

✅ Final Answer:

$$\boxed{x - y = 2n\pi \quad \text{for } n \in \mathbb{Z}}$$