Q219 / 270 · NDA 2012 (Phase 2) · Mathematics

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

  • A
    1 only
  • B
    2 only
  • Correct answer
    C
    Both 1 and 2
  • D
    Neither 1 nor 2

Solution

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

Hence, the correct answer is option: 'C'.

Relevant Questions