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Geometry Level 2

If $\tan^2 x+ \cot^2 x = 2$ , which of the following is the general solution for $x$ ? (Take $n$ to be any integer)

n\pi

n\pi \pm \dfrac{\pi}{3}

n\pi \pm \dfrac{\pi}{4}

None of these choices

n\pi \pm \dfrac{\pi}{6}

n\pi \pm \dfrac{\pi}{2}

1 solution

Akhil Bansal
Dec 6, 2015

$\Rightarrow\large \tan^2x + \cot^2x = 2$ $\Rightarrow \large \tan^2x + \cot^2x - 2\tan x \cot x = 0 \quad \quad [\tan x \cot x = 1 ]$ $\Rightarrow\large (\tan x - \cot x )^2 = 0$ $\Rightarrow\large \tan x = \cot x$ $\Rightarrow\large \tan^2 x = 1$ $\Rightarrow\large \tan x = \pm 1 = \pm \tan\left(\dfrac{\pi}{4}\right)$ $\Rightarrow\large x = n\pi \pm \dfrac{\pi}{4}$

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Simple standard approach.

Nice solution. You could also use the AM-GM inequality to note that $\tan^{2}(x) + \dfrac{1}{\tan^{2}(x)} \ge 2,$ with equality holding iff $\tan^{2}(x) = 1.$

Brian Charlesworth - 5 years, 6 months ago

That's even better...short and sweet..

Akhil Bansal - 5 years, 6 months ago

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