Question

-ln|csc(x)+cot(x)|=ln|csc(x)-cot(x)|

Answer 1

What about it?

Answer 2

Need to prove that both sides are equivalent using assorted identities

Answer 3

Okay, let's manipulate the LHS and try to make the RHS. We know \[-ln(x)=ln(\frac{1}{x})\] So: \[ln(\frac{1}{csc(x)+cot(x)})\] Simplify the denominator:
\[ln(\frac{1}{\frac{1}{sin(x)}+\frac{cos(x)}{sin(x)}}) = ln(\frac{1}{\frac{1 +cos(x)}{sin(x)}})\]

Answer 4

We also know that \[\frac{1}{\frac{a}{b}} = \frac{b}{a}\]So we now have \[ln(\frac{sin(x)}{1+cos(x)})\]Multiply the top and the bottom by 1-cos(x) to get:
\[ln(\frac{sin(x)}{1+cos(x)} * \frac{1-cos(x)}{1-cos(x)}) = ln(\frac{sin(x)(1-cos(x))}{1-cos^2(x)}) = ln(\frac{sin(x)(1-cos(x))}{sin^2(x)})\]
\[ = ln(\frac{(1-cos(x))}{sin(x)}) = ln(\frac{1}{sin(x)}-\frac{cos(x)}{sin(x)}) = ln(csc(x)-cot(x))\]
There's your answer :)