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1.PROVE Sin3x/Sinx-Cos3x/Cosx=2? 2.IF tanA=1/2 , tanB=1/3 and A,B are acute angles, find A+B

1.PROVE Sin3x/Sinx-Cos3x/Cosx=2?
2.IF tanA=1/2 , tanB=1/3 and A,B are acute angles, find A+B

Question Image
Grade:11

1 Answers

Prashant
20 Points
4 years ago
Sol1. We know that Sin3x= 3Sinx-4(Sinx)^3
                       and Cos3x=4(Cosx)^3- 3Cosx
        hence the given expression Sin3x/Sinx-Cos3x/Cosx = (3Sinx-4(Sinx)^3)/Sinx-(4(Cosx)^3- 3Cosx)/Cosx
                                                                                    = 3- 4(Sinx)^2- 4(Cosx)^2 +3
                                                                                    = 6- 4{(Sinx)^2+(Cosx)^2}
                                                                                    = 6- 4 =2          {since (Sinx)^2+(Cosx)^2 =1; an identity}
 
 
Sol2.  tanA=1/2, tanB=1/3  find A+B
           we know that tan(A+B)= (tanA +tanB)/(1- tanA * tanB)
           using above identity we have :
                 tan(A+B)=(1/2 + 1/3)/(1-(1/2 *1/3) = (5/6)/(5/6) =1
                 tan(A+B)=1
                 A+B=arc tan(1)=45 degrees 

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