A) \[\frac{1}{\sqrt{2}}\]
B) \[\sqrt{2}\]
C) \[1\]
D) \[2\]
E) \[\frac{1}{2}\]
Correct Answer: A
Solution :
Let\[u=f(\tan x)\]and\[v=g(\sec x)\] \[\therefore \] \[\frac{du}{dx}=f(\tan x).{{\sec }^{2}}x\] and \[\frac{dv}{dx}=g(\sec x).\sec x\tan x\] \[\therefore \] \[\frac{du}{dv}=\frac{du/dx}{dv/dx}\] \[=\frac{f(\tan x).{{\sec }^{2}}x}{g(\sec x).\sec x\tan x}\] \[=\frac{f(\tan x)\sec x}{g(\sec x)\tan x}\] At \[x=\frac{\pi }{4},\]\[\frac{du}{dv}=\frac{f(1)\sec \frac{\pi }{4}}{g(\sqrt{2})\tan \frac{\pi }{4}}\] \[=\frac{2\times \sqrt{2}}{4\times 1}=\frac{1}{\sqrt{2}}\]
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