# How do you rewrite the expression as a single logarithm and simplify ln(cos^2t)+ln(1+tan^2t)?

Jan 13, 2017

$\ln \left({\cos}^{2} \left(t\right)\right) + \ln \left(1 + {\tan}^{2} \left(t\right)\right) = \textcolor{g r e e n}{0}$

#### Explanation:

Things to remember:

[1]$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{\tan = \frac{\sin}{\cos}}$

[2]color(white)("XXX")color(red)(ln(a)+ln(b)=ln(a * b)

[3]color(white)("XXX")color(red)(cos^2+sin^2=1

[4]$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{\ln \left(a\right) = k \mathmr{if} {e}^{k} = a}$

Therefore
$\ln \left({\cos}^{2} \left(t\right)\right) + \ln \left(1 + {\tan}^{2} \left(t\right)\right)$

$\textcolor{w h i t e}{\text{XXX}} = \ln \left({\cos}^{2} \left(t\right) + {\sin}^{2} \left(t\right)\right)$

$\textcolor{w h i t e}{\text{XXX}} = \ln \left(1\right)$

$\textcolor{w h i t e}{\text{XXX}} = 0$

Jan 13, 2017

$0$.

#### Explanation:

We have to use the Rule $: \ln a + \ln b = \ln \left(a b\right)$, together with the

Identity $: 1 + {\tan}^{2} t = {\sec}^{2} t$

The Exp. $= \ln {\cos}^{2} t + \ln \left(1 + {\tan}^{2} t\right)$

$= \ln {\cos}^{2} t + \ln {\sec}^{2} t$

$= \ln \left\{\left({\cos}^{2} t\right) \left({\sec}^{2} t\right)\right\}$

$= \ln 1$

$= 0$.