if integral 1+tan(x-a)tan(x+a)dx=B ln(cos(x+a)/cos(x-a))+c then find B

1. cot2a
2. tan2a
3. -cot2a
4. -tan2a

1 + tan(x – a) tan(x + a)   =   1 + {sin(x – a)/cos(x – a)} {sin(x + a)/cos(x + a)}  

                                                       ...Using tanA = sinA / cosA

                   = [cos(x + a)cos(x – a) + sin(x + a)sin(x – a)] / [cos(x + a)cos(x – a)]           

                   = cos{(x + a) – (x – a)} / cos(x + a)cos(x – a)  =  cos(2a) / cos(x + a)cos(x – a)

    ...Using cosAcosB + sinAsinB = cos(A – B)

                   = sin(2a)/sin(2a) x cos(2a) / cos(x + a)cos(x – a)

                   = cot(2a) . sin2a / cos(x + a)cos(x – a)  =  cot(2a) sin{(x+a) – (x–a)} / cos(x + a)cos(x – a)

                   = cot(2a) . [sin(x+a)cos(x–a) – cos(x+a)sin(x–a)] / cos(x + a)cos(x – a)

   … Using sin(A – B) = sinAcosB – cosAsinB

                   = cot(2a) [tan(x+a) – tan(x–a)]

∫ [1 + tan(x – a) tan(x + a)]dx  =  ∫cot(2a) [tan(x+a) – tan(x–a)]dx

              =  cot(2a) [∫ tan(x+a) dx – ∫ tan(x–a) dx]    =   cot(2a)[ln(sec(x+a) – ln(sec(x–a))] + c

              =  cot(2a) ln(sec(x + a) / sec(x – a)) + c    =  cot(2a) ln(cos(x – a) / cos(x + a)) + c

              =  cot(2a) [– ln(cos(x + a) / cos(x – a))] + c

              =  – cot(2a) ln(cos(x + a) / cos(x – a)) + c

B = – cot(2a)