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In triangle ABC, if sinAcosB=1/4 and 3tanA=tanB, then cotsquareA is equal to?

In triangle ABC, if sinAcosB=1/4 and 3tanA=tanB, then cotsquareA is equal to?

Grade:11

2 Answers

Debraj Sarkar
17 Points
4 years ago
Cot^2 A=√33tanA = tanB =>3sinAcosB=sinBcosAFrom question sinAcosB=1/4.Putting the value of SinACosB3/4=sinBcosA..............(i)1/4=sinAcosB..............(ii)Dividing (i) by (ii).....=>cotAtanB=3=>cotAcot(90-B)=3=>cotA*cotA=3=>cot^2(A)=3=>cotA=√3
anuridhi
15 Points
2 years ago
         3sinAcosB=sinBcosA
    
 
 
             CosAsinB=3/4
   
 
          Sin(A+B)=1
 
 
 
 
 
 
 
 
C=π/2,B=π/2-A
 
 
 
 
 
3tanA=tan(π/2-A)
 
 
 
 
3tanA=tan(π/2-A)
 
..................................
 
 
3=cot^2A

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