Question

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

• A. $2secx{(secx+tanx)}^2$
• B. $2secx{(secx-tanx)}^2$
• C. $2secx$
• D. None of the above

Answer 1

Answer: (A)

$y=\frac{\sec x+\tan x}{\sec x-\tan x}$
$\Rightarrow$ $y=\frac{\sec x+\tan x}{\sec x-\tan x}\times \frac{\sec x+\tan x}{\sec x+\tan x}$
$\Rightarrow$ $y=\frac{{(\sec x+\tan x)}^2}{{\sec }^{2}x-{\tan }^{2}x}$
$\Rightarrow$ $y=\frac{{(\sec x+\tan x)}^2}{1}$
On differentiating with respect to $x,$
$\frac{dy}{dx}=2(\sec x+\tan x)(\sec x\tan x+{\sec }^{2}x)$
$=2(\sec x+\tan x)\sec x(\tan x+\sec x)$
$=2\sec x{(\sec x+\tan x)}^2$