[qquad [sec cot alpha in(0,pi/2)" be fixed.It the integront "],[int(tan x+tan alpha)/(tan x-tan alpha)dx=A(x)cos2+beta(x)]]

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.

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.

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.

int{1+tan x*tan(x+alpha)}dx=

int tan(x-alpha).tan(x+alpha).tan 2x dx is equal to

int tan(x-alpha).tan(x+alpha).tan 2x dx is equal to

If 2 tan beta + cot beta = tan alpha , prove that 2 tan (alpha-beta) = cot beta

If tan 2alpha = cot 2beta , then sec(alpha + beta) =

if c o* x=cos alpha.cos beta then prove that tan((x+alpha)/(2))*tan((x-alpha)/(2))=(tan^(2)beta)/(2)

(x tan alpha+y cot alpha)(x cot alpha+y tan alpha)-4xy cos^(2)2 alpha=