Grade 12th passTrigonometry
A+B+C=90, ∑ cos(B+C)/cosBcosC how to do this problem
A+B+C=90, ∑ cos(B+C)/cosBcosC
how to do this problem
3 Answers
Aditya Gupta
we have to find out ∑ cos(B+C)/cosBcosC=∑ 1 – sinBsinC/cosBcosC=∑ 1 – tanBtanC=3 – ∑ tanBtanC
now we have A+B+C=90
take tan both sides
(tan A + tan B + tan C - tan A tan B tan C)/( 1 - tan A tan B- tan C tan A - tan B tan C)= not defined, which is possible only if 1 - tan A tan B- tan C tan A - tan B tan C=0
or ∑ tanBtanC=1
so, ∑ cos(B+C)/cosBcosC = 3 – 1
= 2
yojain
we have A+B+C=90
take tan both sides
(tan A + tan B + tan C - tan A tan B tan C)/( 1 - tan A tan B- tan C tan A - tan B tan C)= not defined, which is possible only if 1 - tan A tan B- tan C tan A - tan B tan C=0
or ∑ tanBtanC=1
so, ∑ cos(B+C)/cosBcosC = 3 – 1
= 2
Lalitha Ghanapuram
Prove that Sigma Cos(B+C)÷CosB CosC=2 if A+B+C=π÷2
From the above consider
Sigma Cos(B+C)÷CosB CosC
=Cos(B+C)/CosB CosC + Cos(C+A)/CosC CosA + Cos(A+B)/CosA CosB
=(CosB CosC - SinB SinC)/CosB cosC + (CosC CosA - SinC SinA)/CosC CosA + (CosA CosB - SinA SinB)/CosA CosB
=1 -(SinB SinC/CosB CosC) + 1 -(SinC SinA/CosC CosA) + 1 -(SinA SinB/CosA CosB)
=3 - TanB TanC - TanC TanA - TanA TanB
=3 - (TanB TanC +TanC TanA +TanA TanB) ----(1)
Given that A+B+C = π/2
A+B =π/2 - C
Tan(A+B) =Tan(π/2 - C)
TanA + TanB/(1-TanA TanB) = Cot C
TanA + TanB/(1-TanA TanB) = 1/Tan C
(TanA +TanB)TanC = 1 - TanA TanB
TanA TanC + TanB TanC = 1 - TanA TanB
TanA TanC + TanB TanC + TanA TanB = 1 -----(2)
Put the value of equation (2) in (1),then we get
(1) => Sigma Cos(B+C)÷(CosB CosC) = 3 - 1 = 2
Hence the Proof
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