If tan θ = t, then tan 2θ + sec 2θ = ___________.
Given $\tan \theta=t$
$\tan 2 \theta+\sec 2 \theta$
$=\frac{2 \tan \theta}{1-\tan ^{2} \theta}+\frac{1+\tan ^{2} \theta}{1-\tan ^{2} \theta} \quad\left[\right.$ using identity $\left.\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta} \sec 2 \theta=\frac{1+\tan ^{2} \theta}{1-\tan ^{2} \theta}\right]$
$=\frac{1+\tan ^{2} \theta+2 \tan \theta}{\left(1-\tan ^{2} \theta\right)}$
$=\frac{(1+\tan \theta)^{2}}{1-\tan ^{2} \theta}$
$=\frac{(1+\tan \theta)^{2}}{(1-\tan \theta)(1+\tan \theta)}=\frac{1+\tan \theta}{1-\tan \theta}=\frac{1+t}{1-t}$
therefore, $\tan 2 \theta+\sec 2 \theta=\frac{1+t}{1-t}$
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