Guest

In triangle ABC if cosA cosB cosB=1/3 then find the value of tanAtanB+tanBtanC+tanCtanA

In triangle ABC if cosA cosB cosB=1/3 then find the value of tanAtanB+tanBtanC+tanCtanA

Grade:11

3 Answers

Mayur Reddy
23 Points
6 years ago
 

Given cosAcosBcosC=1/3

then tanAtanB +tanB tanC+ tanC tanA=sinAsinBcosC +sinBsinCcosA +sinCsinAcosB/cosAcosBcosC

tanAtanB +tanB tanC+ tanC tanA=(sinA sinB cosC +sinB sinC cosA +sinC sinA cosB)/cosA cosB cosC

 

                                               =3 (sinA sinB cosC +sinC [sinB cosA + sinA cosB])

 

                                               =3 (sinA sinB cosC +sinC [sin(A+B)])

 

                                               =3 (sinA sinB cosC +sin2C )

 

                                               =3 (sinA sinB cosC +1-cos2C )

 

                                               =3 (1+cosC{-cosC+sinA sinB } )

 

                                               =3 (1+cosC cosA cosB  )

 

                                               =3 (1+1/3 )

                                               =3 (4/3)

                                               =4

Vedit
20 Points
4 years ago
tanAtanB+ tanBtanC+ tanCtanA=1+ secAsecBsecC
SsecAsecBsecC=3
Therefore,tanAtanB+ tanBtanC+ tanCtanA=1+3 =4
Ravi Babu Duvvuru
13 Points
11 months ago
A+B+C=180
B+C=180-A
Sin(B+C)=sin(180-A)
Sin(B+C)= sinA
TanA-TanB-TanC=TanA -[sinB/cosB +sin C/cos C]
Tan A- [sinB cos C+sin C cosB /cosB cos C]
Tan A- sin(B+C)/cosA
TanA -Tan A=0

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free