Solution
Given,
$$\tan x = \cot(45^\circ+ 2x)$$
We know that, $$\tan(90^\circ-\theta)=\cot \theta$$
$$\Rightarrow \tan x = \tan(90^\circ-45^\circ- 2x)$$
$$\Rightarrow \tan x = \tan(45^\circ- 2x)$$
$$\Rightarrow x = 45^\circ- 2x$$
$$\Rightarrow 3x = 45^\circ$$
$$\Rightarrow x = 15^\circ$$
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