Question

Can someone please explain why -ln[cscx+cotx] is the same as ln[cscx-cotx]?

Answer 1

Proof that
\[-\ln|\csc+\cot | = \ln|\csc -\cot |\]

\[LHS\\=-\ln|\csc+\cot|\\
=\ln|\frac1{\csc+\cot}|\\
=\ln|\frac1{\frac1\sin+\frac\cos\sin}|\\
=\ln|\frac\sin{1+\cos}|\\
=\ln|\frac\sin{1+\cos}\times\frac{1-\cos}{1-\cos }|\\
=\ln|\frac{\sin(1-\cos)}{1-\cos^2}|\\
=\ln|\frac{\sin(1-\cos)}{\sin^2}|\\
=\ln|\frac{1-\cos}\sin|\]

\[RHS\\
= \ln|\csc -\cot |\\
= \ln|\frac1\sin-\frac\cos\sin|\\\
= \ln|\frac{1-\cos}\sin|\\
%= \ln|\frac{1-\cos}\sin\times\frac{1+\cos}{1+\cos}|\\
%= \ln|\frac{1-\cos^2}{\sin(1+\cos)}|\\
%= \ln|\frac{\sin^2}{\sin(1+\cos)}|\\
%= \ln|\frac1{1+\cos}|
\]

\[LHS=RHS\qquad \square\]