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`        If sinasinb-cosacosb +1=0 prove that 1+cota tanb=0`
4 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
```
4 years ago
```							sinasinb-cosacosb=-1so,cos(a+b)=-1so,a+b=180 or a=180-b  1+tanbcota1+tanb/tana(tana+tanb)/tana(tan(180-b)+tanb)/tana(-tanb+tanb)/tana=0Hence proved.
```
4 years ago
```							thanks bro
```
4 years ago
```							SinA. SinB - cosA. CosB + 1 = 0-{cos(A+B)} = -1Cos(A+B) =1Let A+B be xThen cosx=1x=0A+B =0ie Sin(A+B)=0= sinAcosB + CosASinB =0Dividing above equation by cosAcosB, we getTanA+ tanB=0TanB= - tanA.        (1) Now we need to prove.. 1+ cotAtanB=0So LHS = 1 + cotA(-tanA)    ( from (1))= 1- cotAtanA=1-1=0=RHS
```
one year ago
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