## Task Condition

Find the derivative.

$y=\frac{1}{\sin \alpha} \ln (\tan x+\cot \alpha)$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{\sin \alpha} \ln (\operatorname{tg} x+\operatorname{ctg} \alpha)\right)^{\prime}=\frac{1}{\sin \alpha} \cdot \frac{1}{\operatorname{tg} x+\operatorname{ctg} \alpha} \cdot \frac{1}{\cos ^{2} x}= \\
& =\frac{1}{\frac{\sin \alpha \cdot \sin x \cdot \cos ^{2} x}{\cos x}+\frac{\sin \alpha \cdot \cos \alpha \cdot \cos ^{2} x}{\sin \alpha}}=\frac{1}{\sin \alpha \cdot \sin x \cdot \cos x+\cos \alpha \cdot \cos ^{2} x}= \\
& =\frac{1}{\cos x \cdot(\sin \alpha \cdot \sin x+\cos \alpha \cdot \cos x)}=\frac{1}{\cos x \cdot \cos (\alpha-x)}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 15-1