ddx(etanx+sinx)=etanx⋅f(x)+g(x), then (g(x))2⋅(f(x))=

ddxetanx+ddxsinx=etanx⋅sec2x+cosx∴f(x)=sec2xg(x)=cosx∴(g(x))2⋅f(x)=cos2x⋅sec2x=1

ddx(etanx+sinx)=etanx⋅f(x)+g(x), then (g(x))2⋅(f(x))=

ddxetanx+ddxsinx=etanx⋅sec2x+cosx∴f(x)=sec2xg(x)=cosx∴(g(x))2⋅f(x)=cos2x⋅sec2x=1

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$\frac{d}{d x} (e^{\tan x} + \sin x) = e^{\tan x} \cdot f (x) + g (x)$ , then ${(g (x))}^{2} \cdot (f (x)) =$

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$- 1$

$= 1$

answer is B.

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Detailed Solution

$\frac{d}{d x} e^{\tan x} + \frac{d}{d x} \sin x = e^{\tan x} \cdot \sec^{2} x + \cos x$

$∴ f (x) = \sec^{2} x$

$g (x) = \cos x$

$∴ {(g (x))}^{2} \cdot f (x)$

$= \cos^{2} x \cdot \sec^{2} x$

$= 1$

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