If sec 2θ + tan 2θ = 5/3, then what is the value of tan2θ?
√3
Given \(\sec^{2}\theta+\tan^{2}\theta=\dfrac{5}{3}\). Using the identity \(\sec^{2}\theta-\tan^{2}\theta=1\):
Add the two equations:
\[2\sec^{2}\theta=\frac{5}{3}+1=\frac{8}{3}\;\Rightarrow\;\sec^{2}\theta=\frac{4}{3}\]
Hence \(\tan^{2}\theta=\sec^{2}\theta-1=\dfrac{1}{3}\), so \(\tan\theta=\dfrac{1}{\sqrt{3}}\).
Double-angle formula:
\[\tan 2\theta=\frac{2\tan\theta}{1-\tan^{2}\theta}=\frac{2/\sqrt{3}}{1-1/3}=\frac{2/\sqrt{3}}{2/3}=\frac{3}{\sqrt{3}}=\sqrt{3}\]
Therefore \(\tan 2\theta=\boldsymbol{\sqrt{3}}\).
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