# If sinasinb-cosacosb +1=0 prove that 1+cota tanb=0

Akshay
185 Points
8 years ago
Hint: use sina*sinb= ½ * (cos(a-b) – cos(a+b)),
and cosa*cosb = ½ * (cos(a-b) + cos(a+b)),
simplify and you will get,  a + b = 2n*pi,
just put in second equation
Akhil Chinnu
45 Points
8 years ago
sinasinb-cosacosb=-1
so,cos(a+b)=-1
so,a+b=180 or a=180-b
1+tanbcota
1+tanb/tana
(tana+tanb)/tana
(tan(180-b)+tanb)/tana
(-tanb+tanb)/tana
=0
Hence proved.
Mohammed Tameem Mohiuddin
10 Points
8 years ago
thanks bro

Tanu
13 Points
5 years ago
SinA. SinB - cosA. CosB + 1 = 0
-{cos(A+B)} = -1
Cos(A+B) =1
Let A+B be x
Then cosx=1
x=0
A+B =0
ie Sin(A+B)=0
= sinAcosB + CosASinB =0
Dividing above equation by cosAcosB, we get
TanA+ tanB=0
TanB= - tanA.        (1)

Now we need to prove.. 1+ cotAtanB=0
So LHS = 1 + cotA(-tanA)    ( from (1))
= 1- cotAtanA
=1-1=0=RHS
Sanjay
13 Points
4 years ago
Sinasinb-cosacosb+1=0
1=cosacosb-sinasinb
cos(a+b)=1
a+b=0,2p,4p....2np
tan(a+b)=tan0
(tana+tanb) /1-tanatanb=0
tana +tanb=0
tanb=-tana ...(1)
now 1+cotatanb=0 from eq.(1)
1+cota(-tana)
=1-1=0