How can I prove $|\tan(x) +\tan(y) | \ge |x+y|$ with the Mean Value Theorem?
By Gabriel Cooper •
How can I prove $$|\tan(x) +\tan(y) | \ge |x+y|$$ for all $x, y\in [-\pi/2,\pi/2]$?
I'm trying to do it with the mean value theorem but I can't use it for this equation, right? Since there is a "+" between $\tan x$ and $\tan y$.
$\endgroup$ 32 Answers
$\begingroup$Apply the Mean Value Theorem to the function $f(x)=\tan(x)$. Since $f'(x)= 1+\tan^{2}(x)\ge1$ we get $$|\tan(x)-\tan(y)|\ge{|x-y|} $$ Note that $$\tan(-y)=-\tan(y).$$ Upon substitution of $-y$ for $y$ in $|\tan(x)-\tan(y)|\ge{|x-y|}$ we get $$|\tan(x)-\tan(-y)|\ge{|x+y|} $$ Thus $$ |\tan(x)+\tan(y)|\ge{|x+y|}.$$
$\endgroup$ $\begingroup$Use the fact that $|\tan(x)-\tan(y)| \geq |x-y| \times \inf({\frac{1}{\cos^2(x)}}) = |x-y|$
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