A) \[\sec x{{a}^{\sec x-\tan x}}\]
B) \[\sin x{{a}^{\tan x-\sec x}}\]
C) \[\sin x{{a}^{\sec x-\tan x}}\]
D) \[{{a}^{\sec x-\tan x}}\]
Correct Answer: C
Solution :
Let\[u={{a}^{\sec x}}\]and\[v={{a}^{\tan x}}\] On differentiating w.r.t.\[x,\]we get \[\frac{du}{dx}={{a}^{\sec x}}{{\log }_{e}}a.\sec x\tan x\] and \[\frac{dv}{dx}={{a}^{\tan x}}{{\log }_{e}}a.{{\sec }^{2}}x\] \[\therefore \] \[\frac{du}{dv}=\frac{{{a}^{\sec x}}{{\log }_{e}}a.\sec x\tan x}{{{a}^{\tan x}}{{\log }_{e}}a.{{\sec }^{2}}x}\] \[={{a}^{\sec x-\tan x}}\sin x\]
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