Question

In a acute triangle
tan A tan B + tan B tan C + tan C tan A > x then x equals
(a) 3
(b) 3^1/2
(c)5
(d) None of these​

Answer 1

Answer:

Step-by-step explanation:

using the property in  acute triangle,  

secAsecBsecC= 1/cosA cos B cos C,

above will be maximum at A=B=C=60  °

 

∴ Max. (cosAcosBcosC)=1/8,

∴ Min. (secAsecBsecC)=8  

Since we have cos(A+B+C)=cosAcosBcosC(1−tanAtanB−tanBtanC−tanCtanA)

Using the above calculated value of angle  A+B+C=π and cosθ=−1,  

putting above values we get,

tanAtanBtanC+tanCtanA=1+secAsecBsecC

∴tanAtanB+tanBtanC+tanCtanA=9

∴tanAtanB+tanBtanC+tanCtanA>x.

x=3.

Hope it helps.