Variation of $x\to \tan x$ over $(\pi/2,\pi ]$ without using derivation?

Question

I'm sorry to ask this trivial question, i have accrossed this problem with my students of high school level , they are not discussed "Derivation method" to set the monotonicity of any real valued function . The main problem is if i assume that : there are two real :$x_1,x_2\in (\pi/2,\pi]$ such that $x_1>x_2$ then we have : $$\sin x_1<\sin x_2\tag{1}$$ , and $$\cos x_1<\cos x_2\tag{2}$$ .

The two function are decreasing in $(\pi/2,\pi]$, The problem it's not allowed to divide the inequality $(1)$ over inequality $(2)$ ., then how do i can get the variation of $x\to \tan x$ using this method or any simple method for this level without derivation?

Answer 1

Note that

$$\sin x>0 \quad \sin x_1<\sin x_2\iff \frac{\sin x_1}{\sin x_2}<1$$

$$\cos x<0 \quad \cos x_1<\cos x_2\iff \frac{\cos x_2}{\cos x_1}<1$$

then since $\tan x<0$

$$\frac{\sin x_1}{\sin x_2}\frac{\cos x_2}{\cos x_1}=\frac{\tan x_1}{\tan x_2}<1\implies \tan x_1>\tan x_2$$

Answer 2

Hint: It might be easier to show that $\tan(x)$ is increasing on $\left[0,\frac\pi2\right)$, then use $\tan(x)=-\tan(\pi-x)$

Answer 3

You have $$sin x_1<sin x_2\tag{1}$$ and $$cos x_1<cos x_2\tag{2}$$ Upon multiplying both sides of $ (2)$ by $-1$ we get $$ -cos x_1>-cos x_2\tag{3}$$ At this point let us compare $ \frac {sin x_1}{-cos x_1}$ with $ \frac {sin x_2}{-cos x_2}$.

Note that for $ x_1,x_2\in (\pi/2,\pi]$ , $ sinx_1< sinx_2$ and $ -cos x_1>-cos x_2 $.

Therefore, $-tanx_1<-tanx_2$ which is equivalent to $tanx_1>tanx_2$.

Thus $tan(x)$ is increasing on the given interval.