Answer
$$\theta=40^\circ$$
Work Step by Step
$$\cot(\theta-10^\circ)=\tan(2\theta-20^\circ)\hspace{1cm}(1)$$
Cofunction identity:
$$\cot\theta=\tan(90^\circ-\theta)$$
So, $$\cot(\theta-10^\circ)=\tan[90^\circ-(\theta-10^\circ)]$$
$$\cot(\theta-10^\circ)=\tan(90^\circ-\theta+10^\circ)$$
$$\cot(\theta-10^\circ)=\tan(100^\circ-\theta)$$
Replace this in $(1)$ for $\cot(\theta-10^\circ)$:
$$\tan(100^\circ-\theta)=\tan(2\theta-20^\circ)$$
$$100^\circ-\theta=2\theta-20^\circ$$
$$3\theta=120^\circ$$
$$\theta=40^\circ$$
