sin (A +B+C)
- sin B
cos(A+B)
If A+B+C = 1, find value of
sin B
0
– tan A
cos C
tan A
0
9a
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Answer:
A
0
Δ=
∣
∣
∣
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sin(A+B+C)
−sinB
cos(A+B)
sinB
0
−tanA
cosC
tanA
0
∣
∣
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
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sin(π)
−sinB
cos(π−C)
sinB
0
−tanA
cosC
tanA
0
∣
∣
∣
∣
∣
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=
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0
−sinB
−cos(C)
sinB
0
−tanA
cosC
tanA
0
∣
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∣
=−sinB(tanAcosC)+cosC(sinBtanA)=0
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