# How can one prove that ln(sec^2x+csc^2x) = ln(sec^2x)+ln(csc^2 x)?

Feb 15, 2017

#### Explanation:

${\sec}^{2} x + {\csc}^{2} x$

= $\frac{1}{\cos} ^ 2 x + \frac{1}{\sin} ^ 2 x$

= $\frac{{\sin}^{2} x + {\cos}^{2} x}{{\sin}^{2} x {\cos}^{2} x}$

= $\frac{1}{{\sin}^{2} x {\cos}^{2} x}$

= ${\sec}^{2} x {\csc}^{2} x$

As ${\sec}^{2} x + {\csc}^{2} x = {\sec}^{2} x {\csc}^{2} x$, taking natural log on both sides

$\ln \left({\sec}^{2} x + {\csc}^{2} x\right) = \ln {\sec}^{2} x + \ln {\csc}^{2} x$