MHB Can You Prove This Trigonometric Inequality?

[anemone]

Prove $\tan 6^{\circ}+\tan 24^{\circ}+\tan 42^{\circ}+\tan 86^{\circ}\gt 4$.

 

[anemone]

Prove $\tan 6^{\circ}+\tan 24^{\circ}+\tan 42^{\circ}+\tan 86^{\circ}\gt 4$.

My solution:

Spoiler

First we know

[TABLE="class: grid, width: 700"]
[TR]
[TD]$\tan x=x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+\cdots$[/TD]
[TD]$86^{\circ}=\dfrac{86^{\circ}\pi}{180^{\circ}}>1.5$ rad[/TD]
[TD]$42^{\circ}=\dfrac{42^{\circ}\pi}{180^{\circ}}>0.733$ rad[/TD]
[/TR]
[/TABLE]

So we have

$\begin{align*}\tan 42^{\circ}+\tan 86^{\circ}&\gt (1.5+\dfrac{1.5^3}{3}+\dfrac{2(1.5)^5}{15}+\cdots)+(0.733+\cdots)\\&>3.6375+0.733\\&>4.3705\end{align*}$

Since $\tan 6^{\circ},\, \tan 24^{\circ}>0$ we can conclude by now $\tan 6^{\circ}+\tan 24^{\circ}+\tan 42^{\circ}+\tan 86^{\circ}\gt 4$ and we're hence done.

 

[kaliprasad]

My solution:

Spoiler

First we know

[TABLE="class: grid, width: 700"]
[TR]
[TD]$\tan x=x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+\cdots$[/TD]
[TD]$86^{\circ}=\dfrac{86^{\circ}\pi}{180^{\circ}}<1.5$ rad[/TD]
[TD]$42^{\circ}=\dfrac{42^{\circ}\pi}{180^{\circ}}<0.733$ rad[/TD]
[/TR]
[/TABLE]

So we have

$\begin{align*}\tan 42^{\circ}+\tan 86^{\circ}&\gt (1.5+\dfrac{1.5^3}{3}+\dfrac{2(1.5)^5}{15}+\cdots)+(0.733+\cdots)\\&>3.6375+0.733\\&>4.3705\end{align*}$

Since $\tan 6^{\circ},\, \tan 24^{\circ}>0$ we can conclude by now $\tan 6^{\circ}+\tan 24^{\circ}+\tan 42^{\circ}+\tan 86^{\circ}\gt 4$ and we're hence done.

should be

Spoiler

$86^{\circ}=\dfrac{86^{\circ}\pi}{180^{\circ}}>1.5$ rad

 

[anemone]

Thanks kaliprasad for catching it, I just fixed the mistakes.