if cosA = tanB , cosB = tanC and cosC = tanA
then numerical value of sinA = ?

• cosA =tanB
  cosA=sinB/cosB
  putting cosB = tanC and cosC = tanA we get
  sinA=(cos^2AsinC)/sinB
• 9 years agoHelpfull: Yes(1) No(2)
• options are like, sqrt5 +- 1 / 2 , sqrt5+-1/4
• 9 years agoHelpfull: Yes(1) No(1)
• sinA =sinC
• 9 years agoHelpfull: Yes(0) No(2)
• sin A/cos A=cos c=>cosc*cos A=sinA
  cos c sinb/cosb=sin A;
  cos c (sinb/sinc)*cos c=sin A;this will be the final eqn,some data is missing to substitute values,maybe you havent given angles
• 9 years agoHelpfull: Yes(0) No(0)
• MANIKANDAN
  
  question is absolutely right and options are
  sqrt5 +- 1 / 2 , sqrt5+-1/4
• 9 years agoHelpfull: Yes(0) No(0)
• sqrt5 +- 1 / 2 is the answer as the sides of triangle are in G.P according to given eq^n .... hence eqn after that will become r^4 + r^2 = 1 putting x= r^2 eqn will be x^2+x-1=0 after solving for x by sridhacharya method we will have answer
  
• 9 years agoHelpfull: Yes(0) No(0)