Question

Consider the following statements:1. If $y=\ln (\sec x+\tan x)$, then $\frac{dy}{dx}=\sec x$.2. If $y=\ln (\mathrm{cosec}x-\cot x)$, then $\frac{dy}{dx}=\mathrm{cosec}x$.Which of the above is / are correct?

• (A) 1 only
• (B) 2 only
• (C) Both 1 and 2
• (D) Neither 1 nor 2

Answer 1

The Correct Option is C

Explanation:
Concept:$y=\ln x$ then $\frac{dy}{dx}=\frac{1}{x}$$y=\tan x$ then $\frac{dy}{dx}={\sec }^{2}x$$y=\sec x$ then $\frac{dy}{dx}=\sec x\cdot \tan x$$y=\mathrm{cosec}x$ then $\frac{dy}{dx}=-\mathrm{cosec}x\cdot \cot x$$y=\cot x$ then $\frac{dy}{dx}=-{\mathrm{cosec}}^{2}x$Let, $y=\ln (\sec x+\tan x)$ $\Rightarrow \frac{dy}{dx}=\frac{1}{\sec x+\tan x}\times (\sec x\cdot \tan x+{\sec }^{2}x)$ $\Rightarrow \frac{dy}{dx}=\frac{\sec x}{\sec x+\tan x}\times (\tan x+\sec x)$ $\Rightarrow \frac{dy}{dx}=\sec x$ Statement (1) is true. Now, $y=\ln (\mathrm{cosec}x-\cot x)$,$\Rightarrow \frac{dy}{dx}=\frac{1}{\mathrm{cosec}x-\cot x}\times (-\mathrm{cosec}x\cdot \cot x+{\mathrm{cosec}}^{2}x)$ $\Rightarrow \frac{dy}{dx}=\frac{\mathrm{cosec}x}{\mathrm{cosec}x-\cot x}\times (\mathrm{cosec}x.-\cot x)$ $\Rightarrow \frac{dy}{dx}=\mathrm{cosec}x$
Statement (2) is true.Hence, the correct option is (C).