Let `alpha in (0, pi//2)` be fixed. If the integral `int("tan x" + "tan" alpha)/("tan x" - "tan" alpha)dx = A(x)"cos 2 alpha+B(x)` `"sin" 2 alpha +C`,where C is a constant of integration, then the functions A(x) and B(x) are respectively.A. `x+alpha "and log"_(e)|"sin"(x+alpha)|`B. `x-alpha "and log"_(e)|"sin"(x-alpha)|`C. `x-alpha "and log"_(e)|"cos"(x-alpha)|`D. `x+alpha "and log"_(e)|"sin"(x-alpha)|`

Let `alpha in (0, pi//2)` be fixed. If the integral `int("tan x" + "ta

1.	Let `alpha in (0, pi//2)` be fixed. If the integral `int("tan x" + "tan" alpha)/("tan x" - "tan" alpha)dx = A(x)"cos 2 alpha+B(x)` `"sin" 2 alpha +C`,where C is a constant of integration, then the functions A(x) and B(x) are respectively.A. `x+alpha "and log"_(e)\|"sin"(x+alpha)\|`B. `x-alpha "and log"_(e)\|"sin"(x-alpha)\|`C. `x-alpha "and log"_(e)\|"cos"(x-alpha)\|`D. `x+alpha "and log"_(e)\|"sin"(x-alpha)\|`
Answer» Correct Answer - B Let `I = int(tan x + tan alpha)/(tan x - tan alpha)dx, alpha in (0,(pi)/(2))` `= int((sinx)/(cosx)+(sin alpha)/(cos alpha))/((sinx)/(cosx)-(sin alpha)/(cos alpha))dx` `=int(sin x cos alpha + sin alpha cos x)/(sin x cos alpha - sin alpha cos x)dx` `=int(sin(x+alpha))/(sin(x-alpha))dx` Now, put `x-alpha = t rArr dx = dt, so` `I=int(sin(t+2 alpha))/(sin t)dt` `=int(sin t cos 2 alpha + sin 2 alpha cos t)/(sin t)dt` `=int(cos 2 alpha + sin 2 alpha (cos t)/(sin t))dt` `= t(cos 2 alpha)+(sin 2 alpha)log_(e)\|sin t\|+C` `=(x-alpha) cos 2 alpha +(sin 2 alpha)log_(e)\|sin(x-alpha)\|+C` `= A(x) cos 2 alpha + B(x) sin 2 alpha + C` (given) Now on comparing, we get `A(x) = x - alpha and B(x) = log_(e)\|sin(x-alpha)\|`

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