In triangle abc angle c is obtuse then which is correcta. tana.tanb <1b. tana.tanb>1c. none

Hi Rishab,

In case that C is obtuse,

Then C>90.

So A+B msut be less than 90... (because in a triangle A+B+C=180, and C is above 90)

So tan(A+B) must be postivie... (Since A+B is acute < 90)

tan(A+B) = (tanA+tanB)/(1-tanAtanB)

So 1-tanAtanB>0..... (Only then tan(A+B) can be positive)

Hence tanAtanB<1

Option (A) would be the answer.

Best Regards,

Ashwin (IIT Madras).

as a+b<90

tan(a+b)<tan90

tana+tanb/1-tana*tanb<infinity

1-tana*tanb>0

tana*tanb<1

so the option is a

In given situation angle c is pbtuse  that means rest two angles a and b are acute (bcoz  sum of trio of angles in a tringle=180)

  finding tana = 1/tanb(by figure where c is obtuse angle at middle) 

 i.e. tana.tanb=1

  so the answer is c. none. 

plz approve my answer if u like