Question

If $f\left(x\right)=\frac{\tan x+\sec x-1}{\tan x-\sec x+1}$ then $f^{\prime }\left(x\right)$ is equal to

• A. $\sec x(\tan x-\sec x)$
• B. $\sec x(\sec x-\tan x)$
• C. $\sec x(\sec x+\tan x)$
• D. None of these

Answer 1

Answer: (C)

$\sec x(\sec x+\tan x)$

$f\left(x\right)=\frac{\tan x+\sec x-1}{\tan x-\sec x+1}$

$=\frac{{\tan }^{2}x+{\sec }^{2}x+2\tan x\sec x-1}{{(1+\tan x)}^2-{\sec }^{2}x}$

$=\frac{{\tan }^{2}x+{\sec }^{2}x+2\tan x\sec x-1}{2\tan x}=\tan x+\sec x$
Thus $f^{\prime }\left(x\right)={\sec }^{2}x+\sec x\tan x=\sec x\left(\sec x+\tan x\right)$