Question

If $y=\frac{\sec x+\tan x}{\sec x-\tan x}$ , then $\frac{dy}{dx}=$

• A. $2\sec x(\sec x+\tan x)$
• B. $2{\sec }^{2}x(\sec x+\tan x{)}^2$
• C. $2\sec x(\sec x+\tan x{)}^2$
• D. $\sec x(\sec x+\tan x{)}^2$

Answer 1

Answer: (C)

$y=\left(\frac{\sec x+\tan x}{\sec x-\tan x}\right)=\frac{1+\sin x}{1-\sin x}$
$\Rightarrow y=\frac{{\left(1+\sin x\right)}^2}{1-{\sin }^{2}x}=\frac{1+{\sin }^{2}x+2\sin x}{{\cos }^{2}x}$
$\Rightarrow y={\sec }^{2}x+{\tan }^{2}x+2\tan x\sec x$
$\Rightarrow y={\left(\sec x+\tan x\right)}^2$
Now Differentiating (i) w.r.t. x, we get ... (i)
$\frac{dy}{dx}=2\left(\sec x+\tan x\right).\left[\sec x\tan x+{\sec }^{2}x\right]$
$=2\sec x\left(\sec x+\tan x\right)\left(\sec x+\tan x\right)$
$=2\sec x{\left(\sec x+\tan x\right)}^2$