SinAsinB-cosAcosB+1=0 then show sin(A+B) and 1+cotAtanB=0
Riyaj Masud , 5 Years ago
Grade 11
Trigonometry> SinAsinB-cosAcosB+1=0 then show sin(A+B) ...
1 Answers
Arun
Last Activity: 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
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