If tan 2θ = cot (θ + 60°), where 2θ and

Given: tan2θ = cot(θ + 60°)

$\tan 2 \theta=\cot \left(\theta+60^{\circ}\right)$

$\Rightarrow \cot \left(90^{\circ}-2 \theta\right)=\cot \left(\theta+60^{\circ}\right) \quad\left(\because \tan \theta=\cot \left(90^{\circ}-\theta\right)\right)$

$\Rightarrow 90^{\circ}-2 \theta=\theta+60^{\circ}$

$\Rightarrow 90^{\circ}-60^{\circ}=\theta+2 \theta$

$\Rightarrow 3 \theta=30^{\circ}$

$\Rightarrow \theta=\frac{30^{\circ}}{3}$

$\Rightarrow \theta=10^{\circ}$

Hence, the value of $\theta$ is $10^{\circ}$.