Given that,
$$\tan 4 \theta = \cot (2\theta + 30^\circ)$$
We know that $$\cot (90-A)=\tan A$$
Hence,
$$\Rightarrow \cot(90^\circ-4\theta)=\cot (2\theta +30^\circ)$$
$$\Rightarrow 90^\circ -4\theta=2\theta + 30^\circ$$
$$\Rightarrow 6\theta=90^\circ -30^\circ$$
$$\Rightarrow \theta=\dfrac{60^\circ}{6}=10^\circ$$
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