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Let a ∈ (0,π/2) be fixed. If the integral \(∫ \frac{tan x + tan α}{tan x - tan α }dx = \)  A(x) cos 2α + B(x) sin 2α + C  , where C is a constant of integration, then the functions A(x) and B(x) are respectively :

(1)   x – α and loge|cos(x – α)|

(2)   x + α and loge|sin(x – α)|

(3)  x + α and loge|sin(x + α)| 

(4)   x – α and loge|sin(x – α)|

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Correct option is (4)   x – α and loge|sin(x – α)|  

Detailed Explaination:

\(∫ \frac{tan x + tan α}{tan x - tan α} dx\) 

=\(∫ \frac{\frac{\sin x }{\cos x}+ \frac{\sin α}{\cos α}}{\frac{\sin x }{\cos x} - \frac{\sin α}{\cos α}} dx\) 

=\(∫ \frac{\frac{\sin x \cos α+ \sin α \cos x}{\cos α .\cos x}}{\frac{\sin x \cos α- \sin α \cos x}{\cos α .\cos x}} dx\) 

=\(\int \frac{\sin x \cos α+ \sin α \cos x}{\sin x \cos α-\sin α \cos x} dx\)  

\(=\int \frac{\sin( x + α )}{\sin( x - α )} dx\) 

\(=\int \frac{\sin( x -α+2α )}{\sin( x - α )} dx\) 

\(=\int \frac{\sin( x -α)\cos2α }{\sin( x - α )} dx+\frac{\cos( x -α)\sin2α }{\sin( x - α )} dx\) 

\(= (x-\alpha )\cos2\alpha+\sin2\alpha \log_e|sin(x-\alpha)| +C\)

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