Trig Functions — Question 9

Question 9

Let $$f(x) = \tan(x) - \sec(x)$$

• Show that $f(x) = \frac{\sin(x) - 1}{\cos(x)}$

• Determine the values of $x \in (0, 2\pi)$ for which $f(x) = 0$

• Analyze the behavior of $f(x)$ near $x = \frac{\pi}{2}$

Question 9 - Solution

(a) Simplify $f(x) = \tan(x) - \sec(x)$:

We write both terms in terms of sine and cosine:

$$\tan(x) = \frac{\sin(x)}{\cos(x)}, \quad \sec(x) = \frac{1}{\cos(x)}$$

Then: $$f(x) = \frac{\sin(x)}{\cos(x)} - \frac{1}{\cos(x)} = \frac{\sin(x) - 1}{\cos(x)}$$

$$\boxed{f(x) = \frac{\sin(x) - 1}{\cos(x)}}$$

(b) Solve $f(x) = 0 \Rightarrow \frac{\sin(x) - 1}{\cos(x)} = 0$

This occurs when the numerator is 0 (and denominator is not 0):

$$\sin(x) - 1 = 0 \Rightarrow \sin(x) = 1 \Rightarrow x = \frac{\pi}{2}$$

But at $x = \frac{\pi}{2}$, $\cos(x) = 0$, so $f(x)$ is undefined.

Therefore, there is no solution to $f(x) = 0$ on $(0, 2\pi)$

$$\boxed{\text{No solution in } (0, 2\pi)}$$

(c) Behavior near $x = \frac{\pi}{2}$

As $x \to \frac{\pi}{2}^{-}$, we have: - $\sin(x) \to 1$ - $\cos(x) \to 0^+$ - So: $f(x) \to \frac{1 - 1}{0^+} = 0$

As $x \to \frac{\pi}{2}^{+}$, we have: - $\sin(x) \to 1$ - $\cos(x) \to 0^-$ - So: $f(x) \to \frac{1 - 1}{0^-} = 0$

But if we take $x$ very close to $\frac{\pi}{2}$, with small deviation, say: $$f(x) = \frac{\sin(x) - 1}{\cos(x)} \Rightarrow \text{Numerator } \approx 0,\ \text{Denominator } \to 0$$

Apply L’Hôpital’s Rule: $$\lim_{x \to \frac{\pi}{2}} f(x) = \lim_{x \to \frac{\pi}{2}} \frac{\cos(x)}{-\sin(x)} = \frac{0}{-1} = 0$$

So: $$\boxed{\lim_{x \to \frac{\pi}{2}} f(x) = 0}$$

But since the function is undefined at $x = \frac{\pi}{2}$, the limit exists but the function is discontinuous there.