Answer
$$\theta=\frac{140^\circ}{3}$$
Work Step by Step
$$\sec\theta=\csc\Big(\frac{\theta}{2}+20^\circ\Big)\hspace{1cm}(1)$$
We already get from the cofunction identity that
$$\sec\theta=\csc(90^\circ-\theta)$$
That means using that in $(1)$, it can lead to
$$\csc(90^\circ-\theta)=\csc\Big(\frac{\theta}{2}+20^\circ\Big)$$
$$90^\circ-\theta=\frac{\theta}{2}+20^\circ$$
$$\frac{3\theta}{2}=70^\circ$$
$$\theta=\frac{70^\circ}{\frac{3}{2}}=\frac{70^\circ\times2}{3}$$
$$\theta=\frac{140^\circ}{3}$$
