Consider the following statements : 1. If y= ( x+ x) , then (d...

$$\begin{aligned} & y=\log _e(\sec x+\tan x) \\ & \therefore \quad \frac{d y}{d x}=\frac{1}{\sec x+\tan x} \cdot \frac{d}{d x}(\sec x+\tan x)\end{aligned}$$$$\begin{aligned} & =\frac{1}{\sec x+\tan x} \times\left(\sec x \cdot \tan x+\sec ^2 x\right) \\ & =\frac{(\sec x+\tan x) \sec x}{(\sec x+\tan x)}=\sec x\end{aligned}$$y=\log _\rho(\operatorname{cosec} x-\cot x)\frac{d y}{d x}=\frac{1}{\operatorname{cosec} x-\cot x} \cdot \frac{d}{d x}(\operatorname{cosec} x-\cot x)$$\begin{aligned} & =\frac{1}{\operatorname{cosec} x-\cot x} \times\left(-\operatorname{cosec} x \cdot \cot x+\operatorname{cosec}^2 x\right) \\ & =\frac{\operatorname{cosec} x(\operatorname{cosec} x-\cot x)}{(\operatorname{cosec} x-\cot x)}=\operatorname{cosec} x\end{aligned}$$\therefore$ Both 1 and 2 are correct.