Solution
$$\cosec 3\theta = \sec(20^\circ + 2\theta)$$
We know that $$\cosec (90^\circ-\theta)=\sec \theta$$
$$\Rightarrow \cosec 3\theta = \cosec(90^\circ - 20^\circ - 2\theta)$$
$$\Rightarrow \cosec 3\theta = \cosec(70^\circ - 2\theta)$$
From the above,
$$\Rightarrow 3\theta=70^\circ-2\theta$$
$$\Rightarrow 5\theta=70^\circ$$
$$\Rightarrow \theta=14^\circ$$
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