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

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Answer 1 · 2017-02-15T12:30:20

Please see below.

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

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

= $\frac{{sin}^{2} x + {cos}^{2} x}{{sin}^{2} x {cos}^{2} x}$ $(sin^2x+cos^2x)/(sin^2xcos^2x)$

= $\frac{1}{{sin}^{2} x {cos}^{2} x}$ $1/(sin^2xcos^2x)$

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

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

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

How can one prove that ln(sec2x+csc2x)=ln(sec2x)+ln(csc2x)ln(sec^2x+csc^2x) = ln(sec^2x)+ln(csc^2 x)?