The solution set of the equation tan (pi tan | vedclass

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Question

(C) Given the equation: $	an (\pi 	an x) = \cot (\pi \cot x)$ Using the identity $\cot 	heta = 	an (\frac{\pi}{2} - 	heta)$,we get: $	an (\pi

Vedclass · Accepted Answer

$\phi$

Vedclass · Answer

(C) Given the equation: $	an (\pi 	an x) = \cot (\pi \cot x)$ Using the identity $\cot 	heta = 	an (\frac{\pi}{2} - 	heta)$,we get: $	an (\pi 	an x) = 	an (\frac{\pi}{2} - \pi \cot x)$ This implies $\pi 	an x = n\pi + (\frac{\pi}{2} - \pi \cot x)$ for some integer $n$. For simplicity,consider the principal case $n=0$: $\pi 	an x = \frac{\pi}{2} - \pi \cot x$ $	an x + \cot x = \frac{1}{2}$ $\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{1}{2}$ $\frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{2}$ $\frac{1}{\frac{1}{2} \sin 2x} = \frac{1}{2}$ $\frac{2}{\sin 2x} = \frac{1}{2}$ $\sin 2x = 4$ Since the range of $\sin 2x$ is $[-1, 1]$,there is no real value of $x$ that satisfies $\sin 2x = 4$. Thus,the solution set is $\phi$.

The solution set of the equation $\tan (\pi \tan x) = \cot (\pi \cot x)$ for $x \in (0, \frac{\pi}{2})$ is

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