Question

The trigonometric function y = tan x in the II quadrant

• A. decrease from 0 to \(\infty\)
• B. increases from 0 to \(\infty\)
• C. decrease from -\(\infty\) to 0
• D. increases from -\(\infty\) to 0

Answer 1

The Correct Option is D

The function \( y = \tan x \) is defined as: \[ \tan x = \frac{\sin x}{\cos x} \] In the II quadrant (where \( \frac{\pi}{2} < x < \pi \)):

• \(\sin x\) is positive
• \(\cos x\) is negative

Hence, \(\tan x = \frac{+}{-} = -\) ⇒ \( \tan x < 0 \) As \(x\) increases from \(\frac{\pi}{2}^+\) to \(\pi^-\), \(\cos x\) increases (less negative to 0), so \(\tan x\) becomes less negative (i.e., increases from \(-\infty\) to 0).

 

Correct Answer: increases from -\(\infty\) to 0

Answer 2

We need to analyze the behavior of the trigonometric function \(y = \tan x\) in the second quadrant (Quadrant II).

The second quadrant corresponds to angles \(x\) such that \(\frac{\pi}{2} < x < \pi\) (or 90° < x < 180°).v

Let's examine the values of \(\tan x\) at the boundaries and within this quadrant:

• As \(x\) approaches \(\frac{\pi}{2}\) from the right side (i.e., \(x \to \frac{\pi}{2}^+\)), \(\sin x\) approaches 1, and \(\cos x\) approaches 0 from the negative side (\(\cos x < 0\) in QII). Therefore, \(\tan x = \frac{\sin x}{\cos x}\) approaches \(\frac{1}{0^-}\), which tends to \(-\infty\).
• As \(x\) approaches \(\pi\) from the left side (i.e., \(x \to \pi^-\)), \(\sin x\) approaches 0 (\(\sin x > 0\) in QII), and \(\cos x\) approaches -1. Therefore, \(\tan x = \frac{\sin x}{\cos x}\) approaches \(\frac{0}{-1}\), which is 0.

So, as \(x\) increases through the second quadrant (from \(\frac{\pi}{2}\) to \(\pi\)), the value of \(\tan x\) changes from \(-\infty\) to 0.

Since the value is moving from very large negative numbers towards 0, the function is increasing.

Thus, in the second quadrant, \(y = \tan x\) increases from \(-\infty\) to 0.

Comparing this with the given options, the correct option is:

increases from \(-\infty\) to 0