Question

The solution of $e^x\tan ydx+(1-e^x){\sec }^{2}ydy=0$ is

• A. $\tan y=c(1-e^x)$
• B. $\sec y=c(1-e^x)$
• C. $\tan y(1-e^x)=c$
• D. $\sec y=1-e^x$

Answer 1

Answer: (A)

$\tan y=c(1-e^x)$

$\frac{-e^{x}dx}{-e^x+1}=\frac{{\sec }^{2}ydy}{\tan y};\left[\begin{array}{cc}put-(e^x-1)=K;& \tan y=t\\ \Rightarrow -e^{x}dx=dK& {\sec }^{2}ydy=dt\end{array}\right]$
$\Rightarrow \log (e^x-1)+\log c=\log tany$
$\Rightarrow -c(e^x-1)=\tan y$
$\Rightarrow \tan y=c(1-e^x)$