i was wondering if anyone could tell me how tan(lgx/y)+tan(lgy/z)+tan(lgz/x)=tan(lgx/y)tan(lgy/z)tan(lgz/x)?

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we have to prove that tan(log(x/y) + tan(log(y/z)) + tan(log(z/x) = tan(log(x/y))tan(log(y/z))tan(log(z/x))

we have to note that log(x/y) = logx - logy = A-B,

log(y/z) = logy - logz= B-C

log(z/x) = logz- logx = C- A,....................(I)

(to avoid cofusion let us suppose that logx= A, logy=B, logz = C),

apply tan on both sides of equations of (I)

such that tan(log(x/y)) = tan(logx-logy) = tan(A-B)

tan(log(y/z) = tan(logy - logz) = tan(B-C)

tan(log(z/x)) = tan(logz - logx) = tan(C- A)

to prove the question let us prove that tan(A- B)+ tan(B-C) +tan(c-A) = tan(A-B).tan(B-C).tan(C-A)

step 1:

we know that (A-B)+ (B-C) + (C-A) = 0

from this we say that (A-B) + (B-C) = -(C-A), apply tan on both sides

tan((A-B) + (B-C)) = tan(-(C-A)) = -tan(C-A)....................(1)

we have thae formula tan(X+Y) = (tanX +tanY)/(1-tanX.tanY),apply this formula to equation (1)

(tan(A-B) + tan(B-C)) / (1-tan(A-B).tan(B-C) )= -tan(C-A),the denominaor multiplies across

tan(A-B) +tan(B-C) = -tan(C-A) ( 1-tan(A-B).tan(B-C))

= -tan(C-A) + tan(A-B).tan(B-C).tan(C-A)..........(2)(by the distributive law of multiplication)

add tan(C-A) on both sides of equation (2)

thus we get tan(A-B) + tan(B-C) + tan(C-A) = -tan(C-A) + tan(C-A) +tan(A-B).tan(B-C).tan(C-A) = tan(A-B).tan(B-C).tan(C-A)

that is tan(A-B) + tan(B-C) + tan(C-A) = tan(A-B).tan(B-C).tan(C-A)

plug in back the values of A-B,B-C,C-A from ..................(I)

we get tan(logx/y)) + tan(log(y/z))+tan(log(z/x)) = tan(log(x/y)).tan(log(y/z)).tan(log(z/x)), thus the proof.

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