If A + B = Π Prove that tanA + tanB + tan C = tanA. tanB. tanC
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Answer:
There is an error in the question. It should be A+B+C=π.
Step-by-step explanation:
If A+B+C=π, then
tan(A+B+C)=tan(π)
=> tan(A+B+C)= 0
=>(tanA+ tanB + tanC - tanA.tanB.tanC)/ (1-tanA.tanB - tanA.tanC - tanB.tanC)=0
=>tanA+ tanB + tanC - tanA.tanB.tanC=0
=>tanA+ tanB + tanC = tanA.tanB.tanC.
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