If $$\cos \theta = \frac{2p}{1 + p^2}$$, then $$\tan \theta$$ is equal to:
Given that,
$$\cos \theta = \frac{2p}{1 + p^2}$$
$$\sin \theta=\sqrt{1-\cos^2\theta}=\sqrt{1-(\frac{2p}{1 + p^2})^2}=\sqrt{\dfrac{(1+p^2)^2-4p^2}{(1+p^2)^2}}$$
$$\sin \theta=\sqrt{\dfrac{1+p^4+2p^2-4p^2}{(1+p^2)^2}}=\sqrt{\dfrac{1+p^4-2p^2}{(1+p^2)^2}}=\sqrt{\dfrac{(1-p^2)^2}{(1+p^2)^2}}=\dfrac{1-p^2}{1+p^2}$$
So, $$\tan \theta=\dfrac{\dfrac{1-p^2}{1+p^2}}{ \frac{2p}{1 + p^2}}=\dfrac{1-p^2}{2p}$$
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