if a+b+c=180^0 then (cot b + cot c) (cot a + cot b) (cot c + cot a) =
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Given : A+B+C=180
o
........ (i)
Let us solve (cotB+cotC)(cotC+cotA)(cotA+cotB)
=(
sinB
cosB
+
sinC
cosC
)(
sinC
cosC
+
sinA
cosA
)(
sinA
cosA
+
sinB
cosB
)
=(
sinBsinC
cosBsinC+sinBcosC
)(
sinCsinA
cosCsinA+sinCcosA
)(
sinAsinB
cosAsinB+sinAcosB
)
=(
sinBsinC
sin(B+C)
)(
sinCsinA
sin(A+C)
)(
sinAsinB
sin(A+B)
)
=(
sinBsinC
sin(180
o
−A)
)(
sinCsinA
sin(180
o
−B)
)(
sinAsinB
sin(180
o
−C)
) ......... [Using (i)]
=(
sinBsinC
sinA
)(
sinCsinA
sinB
)(
sinAsinB
sinC
) ........ [∵sin(180
o
−x)=sinx]
=
sinAsinBsinC
1
=cscAcscBcscC
Hence, (cotB+cotC)(cotC+cotA)(cotA+cotB)=cscAcscBcscC
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