Question

If tan θ = t, then tan 2θ + sec 2θ = ___________.

Answer 1

$$\begin{gathered} \mathrm{Given}\tan \theta =t \\ \tan 2\theta +sec2\theta \\ =\frac{2\tan \theta }{1-{\tan }^2\theta }+\frac{1+{\tan }^2\theta }{1-{\tan }^2\theta }\left[\mathrm{using}\mathrm{identity}\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }\sec 2\theta =\frac{1+{\tan }^2\theta }{1-{\tan }^2\theta }\right] \\ =\frac{1+{\tan }^2\theta +2\tan \theta }{\left(1-{\tan }^2\theta \right)} \\ =\frac{{\left(1+\tan \theta \right)}^2}{1-{\tan }^2\theta } \\ =\frac{{\left(1+\tan \theta \right)}^2}{\left(1-\tan \theta \right)(1+\tan \theta )}=\frac{1+\tan \theta }{1-\tan \theta }=\frac{1+t}{1-t} \\ \mathrm{therefore},\tan 2\theta +\sec 2\theta =\frac{1+t}{1-t} \end{gathered}$$