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Geometry Level 2

How many real $x$ satisfy $\cot^{-1}(\cot x)= \cot ( \cot^{-1}(x))$ ?

1 2 There's infinitely many solution 0

1 solution

Akhil Bansal
Sep 19, 2015

$\cot^{-1}(\cot x) = x \quad \quad \left(\dfrac{-\pi}{2} , \dfrac{\pi}{2} \right)$

$\cot(\cot^{-1}x) = x \quad \quad \left(\dfrac{-\pi}{2} , \dfrac{\pi}{2} \right)$

Therefore,
$\cot(\cot^{-1}x) = \cot^{-1}(\cot x) = x \quad \left(\dfrac{-\pi}{2} , \dfrac{\pi}{2}\right)$

And,there are $\infty$ numbers lies between $\left(\dfrac{-\pi}{2} , \dfrac{\pi}{2} \right)$

All the values from (0,pi) satisfy cot^-1(cot x)=cot(cot^-1 x), the graph of cot^-1(cot x) can be drawn by splitting cases for x and plotting (its periodic) (or just use desmos graphing calculator) and the graph of cot(cot^-1 x) is just y=x because cot(cot^-1 x)=x for all real values of x. Now these two graphs completely coincide in the region (0,pi)

INFO WEB - 8 months, 2 weeks ago

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