91Ó°ÊÓ

We know that \(ln(a) + ln(b) = ln(a \times b)\). Using this property, the expression can be rewritten as: \(ln(|\cot t|\cdot (1+ \tan^2 t))\).

Knowing \(\tan t = \frac{\sin t}{\cos t}\) and \(\cot t = \frac{1}{\tan t} =\frac{\cos t}{\sin t}\), the expression inside the logarithm is replaced as: \(ln\left(\frac{\cos t|\cos t|}{\sin t}(1+ \frac{\sin^2 t}{\cos^2 t})\right)\).

To simplify it, distribute the fraction over the sum: \(ln\left(\frac{\cos t|\cos t|}{\sin t} + \frac{\cos t|\cos t|\sin^2 t}{\sin t\cos^2 t}\right)\). Then, \(\frac{\cos t|\cos t|\sin^2 t}{\sin t\cos^2 t}\) simplifies to \(\frac{\sin t|\cos t|}{\cos t}\) and the expression becomes: \(ln\left(\frac{\cos t|\cos t|}{\sin t} + \frac{\sin t|\cos t|}{\cos t}\right)\). Next, find a common denominator and add the fractions: \(ln\left(\frac{\cos^2 t|\cos t|+ \sin^2 t|\cos t|}{\cos t \sin t}\right)\). This simplifies to \(ln\left(\frac{|\cos t|}{\cos t\sin t}\right)\). The expression \(cos^2 t + sin^2 t=1\), so that leaves us with: \(ln\left(\frac{1}{\cos t\sin t}\right)\). The reciprocal of the fraction inside the logarithm is brought up to the top as an exponent: \(ln\left((\cos t\sin t)^{-1}\right)\). The expression can be further simplified by replacing \(cos t sin t\) with \(\frac{1}{2}\sin(2t)\) due to double angle identity, resulting in: \(ln\left((\frac{1}{2}\sin 2t)^{-1}\right)\). Finally, rearrange the expression using the property \(ln(a^{-b})=-bln(a)\), the final version of the expression will be: \(-ln(\frac{1}{2}\sin 2t)\).