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I would like to check whether $f: \Bbb [-\pi/2, \pi/2] \in \Bbb R \rightarrow \Bbb R$ where $f(x) = \tan x$ is monotone increasing function.

Of course it looks fairly true in my bare eye with denoted in the printed version of textbook, however, want to show it from the ground.

Which approach is a valid sense to show the monotone property in this case and general case also?

Daschin
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  • Have you tried to check whether its prime derivative is positive? – Matheman Jun 10 '17 at 08:40
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    Actually, $\tan$ is not a function with domain $\left[-\frac\pi2,\frac\pi2\right]$ and codomain $\Bbb R$ to begin with. –  Jun 10 '17 at 08:40

2 Answers2

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HINT: $f'(x)=\frac{1}{\cos^2(x)}>0$ for all $x$ from the given interval we use the Quotient rule: $$\tan(x))'=\frac{\sin^2(x)+\cos^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}>0$$

Dr. Sonnhard Graubner
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  • hm.. ${d\over dx}\tan x={d\over dx}{\sin x \over \cos x} = {d\over dx}(-1){-\sin x \over \cos x}= {d\over dx}(-1){1\over t}{dt \over dx}$ where $t = \cos x$ I think I almost had forgotten how to do differentiate with a parameters.. any help? – Daschin Jun 10 '17 at 08:50
  • ah.. it looks better to use a quotient rule.. – Daschin Jun 10 '17 at 08:54
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The domain should be $(-\pi/2,\pi/2)$.

Since $\displaystyle f'(x)=\sec^2x>0$ for all $\displaystyle x\in\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$, $f( x)$ is (strictly) increasing.

Alternatively, for $\pi/2\le x\le y\le\pi/2$,

\begin{align} f(y)-f(x)&=\frac{\sin y}{\cos y}-\frac{\sin x}{\cos x}\\ &=\frac{\sin y\cos x-\cos y\sin x}{\cos y\cos x}\\ &=\frac{\sin(y-x)}{\cos y\cos x}\\ \end{align}

Note that $\displaystyle y-x\in(0,\pi)$. So, we have $\sin(y-x)>0$.

Since $\cos y\cos x>0$, $f(y)>f(x)$.

$f$ is (strictly) increasing.

CY Aries
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