If tan θ = t, Then Find tan 2θ + sec 2θ

Question:

\[ \text{If } \tan\theta=t, \text{ then find } \tan 2\theta+\sec 2\theta. \]

Solution

Using the identities

\[ \tan 2\theta=\frac{2t}{1-t^2} \] and \[ \sec 2\theta=\frac{1+t^2}{1-t^2} \]

Therefore,

\[ \tan 2\theta+\sec 2\theta = \frac{2t}{1-t^2} + \frac{1+t^2}{1-t^2} \] \[ = \frac{1+2t+t^2}{1-t^2} \] \[ = \frac{(1+t)^2}{(1-t)(1+t)} \] \[ = \frac{1+t}{1-t} \]

Answer

\[ \boxed{\frac{1+t}{1-t}} \]