If sin α*sin β -cos α*cosβ =1 , show that tan α +tan β = 0

Question

If sin α*sin β -cos α*cosβ =1 , show that tan α +tan β = 0

Answer 1

Yes,  I agree.

sinAsinB - cosAcosB = 1 <--- given

cosAcosB - sinAsinB = -1 <--- multiplies both sides by -1

cos(A+B) = -1 <--- angle addition formula for cosine

A+B is an odd multiple of pi   <--- cosine function is -1 at 180=pi, 540 = 3*pi, 900=*5*pi, etc

sin(A+B)=0 <-- sine function is zero at odd multiples of pi

sinAcosB + sinBcosA = 0 <--- angle addition formula for sine

sinA/cosA + sinB/cosB = 0 <--- divides both sides by cosA*cosB

tanA + tanB = 0 <--- tan = sin/cos

Answer 2

sin α*sin β -cos α*cosβ =1 ⇒cos (α+β)=-1.

1+tg²(α+β)=1/cos²(α+β)=1 ⇒ 1+tg²(α+β)=1 ⇒tg(α+β)=0.

tan (α +β) =(tan α +tan β )/(1- tan α tan β) =0

 ⇒ tan α +tan β=0