Question

If $\tan \mathrm{\theta }=t$ then $\tan 2\mathrm{\theta }+\sec 2\mathrm{\theta }=$

(a) $\frac{1+t}{1-t}$

(b) $\frac{1-t}{1+t}$

(c) $\frac{2t}{1-t}$

(d) $\frac{2t}{1+t}$

Answer 1

(a) $\frac{1+t}{1-t}$

$$\begin{gathered} \tan 2\theta +\sec 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }+\frac{1+{\tan }^2\theta }{1-{\tan }^2\theta } \\ =\frac{2\tan \theta +1+{\tan }^2\theta }{1-{\tan }^2\theta } \\ =\frac{{\left(1+\tan \theta \right)}^2}{1-{\tan }^2\theta } \\ =\frac{\left(1+\tan \theta \right)\left(1+\tan \theta \right)}{\left(1+\tan \theta \right)\left(1-\tan \theta \right)} \\ =\frac{1+\tan \theta }{1-\tan \theta } \\ =\frac{1+t}{1-t}\left[\tan \theta =t\left(\mathrm{given}\right)\right] \end{gathered}$$