Question

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

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