How can one prove that #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^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#