If $f(x) = \begin{cases} \frac{\tan(\frac{\pi}{4} - x)}{\cot 2x} & , x ≠ \frac{\pi}{4} \\ 2K + 1 & , x = \frac{\pi}{4} \end{cases}$ is continuous at $x = \frac{\pi}{4}$, then the value of K is equal to

For $f(x) = \begin{cases} \frac{\tan(\frac{\pi}{4} - x)}{\cot 2x} & , x ≠ \frac{\pi}{4} \\ 2K + 1 & , x = \frac{\pi}{4} \end{cases}$ to be continuous at $x = \frac{\pi}{4}$:

$\lim_{x \to \frac{\pi}{4}} \frac{\tan(\frac{\pi}{4} - x)}{\cot 2x} = 2K + 1$

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Substituting $h = x - \frac{\pi}{4}$:

When $x \to \frac{\pi}{4}$, then $h \to 0$

$\frac{\pi}{4} - x = -h$

$x = \frac{\pi}{4} + h$

The limit becomes:

$\lim_{h \to 0} \frac{\tan(-h)}{\cot(2(\frac{\pi}{4} + h))}$

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Using $\tan(-h) = -\tan h$:

$\cot(2 \times \frac{\pi}{4} + 2h) = \cot(\frac{\pi}{2} + 2h)$

Using $\cot(\frac{\pi}{2} + \theta) = -\tan(\theta)$:

$\cot(\frac{\pi}{2} + 2h) = -\tan(2h)$

The limit becomes:

$\lim_{h \to 0} \frac{-\tan h}{-\tan(2h)}$

$= \lim_{h \to 0} \frac{\tan h}{\tan(2h)}$

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Using $\tan(2h) = \frac{2\tan h}{1 - \tan^2 h}$:

$\lim_{h \to 0} \frac{\tan h}{\frac{2\tan h}{1 - \tan^2 h}}$

$= \lim_{h \to 0} \frac{\tan h \times (1 - \tan^2 h)}{2\tan h}$

$= \lim_{h \to 0} \frac{1 - \tan^2 h}{2}$

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As $h \to 0$, $\tan h \to 0$:

$\frac{1 - 0^2}{2} = \frac{1}{2}$

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For continuity:

$2K + 1 = \frac{1}{2}$

$2K = \frac{1}{2} - 1$

$2K = -\frac{1}{2}$

$K = -\frac{1}{4}$

Therefore, $K = -\frac{1}{4}$.