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

Question

1480 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-15T12:30:20

Please see below.

$sec^2x+csc^2x$

= $1/cos^2x+1/sin^2x$

= $(sin^2x+cos^2x)/(sin^2xcos^2x)$

= $1/(sin^2xcos^2x)$

= $sec^2xcsc^2x$

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

$ln(sec^2x+csc^2x)=lnsec^2x+lncsc^2x$

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