If $$\tan 4\theta = \cot(40^\circ - 2\theta), then \theta$$ is equal to:
As per the question,
$$\tan 4\theta = \cot(40^\circ - 2\theta)$$
We know that $$\cot \theta=\tan(90^\circ- \theta)$$
So, $$\tan 4\theta = \tan(90^\circ -40^\circ + 2\theta)$$
$$\tan 4\theta = \tan(50^\circ + 2\theta)$$
Hence, $$4\theta =50^\circ +2 \theta$$
$$\Rightarrow 2\theta =50^\circ $$
$$\Rightarrow \theta =25^\circ $$
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