Why does $\ln(2+\tan\theta) - \frac{1}{2}\ln(1+\tan^2\theta) =\ln (\frac{2 + \tan\theta}{\sec\theta})$?

Question

How is it that

$$\ln(2+\tan\theta) - \frac{1}{2}\ln\left(1+\tan^2\theta\right) = \ln \left(\frac{2 + \tan\theta}{\sec\theta}\right)$$

Answer 1

$$\ln(2+\tan\theta) - \frac{1}{2}\ln\left(1+\tan^2\theta\right) = \ln \left(\frac{2 + \tan\theta}{\sec\theta}\right)$$

$$\ln(2+\tan\theta) - \ln\left(\sec \theta\right ) = \ln \left(\frac{2 + \tan\theta}{\sec\theta}\right)$$

Answer 2

Hint: $$1+\tan^2\theta=\sec^2\theta$$ and $$\ln x- \frac{1}{2}\ln y=\ln x-\ln y^{\frac{1}{2}}=\ln\frac{x}{y^\frac{1}{2}}.$$ I think you can finish from here.

Answer 3

$\begin{align}\ln(1+\tan\theta) - \frac{1}{2}\ln(1+\tan^2\theta) = & \ln \left(\frac{2 + \tan\theta}{\sec\theta}\right)\\= &\ln\left(\frac{(1+\tan\theta)}{\sqrt{1+\tan^2\theta}}\right)\\= &\ln\left(\frac{(1+\tan\theta)}{\sqrt{\sec^2\theta}}\right)\\= &\ln \left(\frac{1 + \tan\theta}{\lvert\sec\theta\rvert}\right) \end{align}$

Answer 4

Hint: Forget $\sec \theta$ (it's just $\dfrac1{\cos\theta}$) and remember $1+\tan^2\theta=\dfrac1{\cos^2\theta}.$