If ∫[1+tan⁡xtan⁡(x+α)]dx=Alog⁡sec⁡(x+α)f(x)+c then

I=∫[1+tan⁡x⋅tan⁡(x+α)]dxI=∫cos⁡xcos⁡(x+α)+sin⁡x⋅sin⁡(x+α)cos⁡x⋅cos⁡x+αdx=∫cos⁡(x+α−x)cos⁡x⋅cos⁡(x+α)dx =cosαsinα∫sinx+α−xcos⁡x⋅cos⁡(x+α)dx=cot⁡α∫sin⁡(x+α)cos⁡(x+α)dx−∫sin⁡xcos⁡xdx=cot⁡α∫tanx+α−∫tanx=cot⁡αlogsecx+αsecx+cA=cot⁡α;f(x)=sec⁡x

If ∫[1+tan⁡xtan⁡(x+α)]dx=Alog⁡sec⁡(x+α)f(x)+c then

I=∫[1+tan⁡x⋅tan⁡(x+α)]dxI=∫cos⁡xcos⁡(x+α)+sin⁡x⋅sin⁡(x+α)cos⁡x⋅cos⁡x+αdx=∫cos⁡(x+α−x)cos⁡x⋅cos⁡(x+α)dx =cosαsinα∫sinx+α−xcos⁡x⋅cos⁡(x+α)dx=cot⁡α∫sin⁡(x+α)cos⁡(x+α)dx−∫sin⁡xcos⁡xdx=cot⁡α∫tanx+α−∫tanx=cot⁡αlogsecx+αsecx+cA=cot⁡α;f(x)=sec⁡x

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If $\int [1 + \tan xtan (x + α)] dx = Alog |\frac{\sec (x + α)}{f (x)}| + c$ then

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$A = \cot α$

$A = \tan α$

f(x) = sec x

f(x) = cos x

answer is A, C.

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Detailed Solution

$\begin{array}{l} I = \int [1 + \tan x \cdot \tan (x + α)] dx \\ I = \int \frac{\cos xcos (x + α) + \sin x \cdot \sin (x + α)}{\cos x \cdot \cos (x + α)} dx \\ = \int \frac{\cos (x + α - x)}{\cos x \cdot \cos (x + α)} dx = \frac{\cos α}{\sin α} \int \frac{\sin (x + α - x)}{\cos x \cdot \cos (x + α)} dx \\ = \cot α [\int \frac{\sin (x + α)}{\cos (x + α)} dx - \int \frac{\sin x}{\cos x} d x] \\ = \cot α [\int \tan (x + α) - \int \tan x] \\ = \cot α [\log \frac{\sec (x + α)}{secx}] + c \\ A = \cot α; f (x) = \sec x \end{array}$