Physics, asked by geethanjali629, 7 months ago

if a+b+c=180^0 then (cot b + cot c) (cot a + cot b) (cot c + cot a) =​

Answers

Answered by brindaMS
1

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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