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Prove tan(a+b)

Q 5 (a) prove that     tan(A+B) =    tan A+tan B
                                                   __________
                                                   1- tan A tan B  
     
(b) show that value of 75 degrees = 2+root3
     
(c) is hence or otherwise show tan(15 degrees) in surd form


I know how to do part a its only a proof but could you show me how to do (b) and (c) thanks
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2 Answers

Prove : Tan ( A+ B ) = (Tan A + Tan B ) / ( 1 - Tan A Tan B) :
 
      Tan ( A + B ) = Sin ( A + B) / Cos ( A +B)
 
                               ( Sin A Cos B + Sin B Cos A ) / ( Cos A Co B - Sin A Sin B)  ( 1)
 
                                Divide both sides of equation ( 1) by Cos A . Cos  B
 
               ( Sin A Cos B + Sin B Cos A )/ (CosA Cos  B )  / right side of ( 1)
 
                         Tan A + Tan B ( i)
 
                   Cos A Cos B - Sin A Sin B / Cos A . Cos B =
                  
                      ( 1 - Tan A Tan B )  ( ii )
 
            Therefore:
   
          Tan ( A + B ) =  ( i) / ( ii )
 
                             = ( Tan A + Tan B )/ ( 1 - Tan A Tan B)
         
                      
  Tan ( 75) = (Tan 45° + Tan 30°) / ( 1 - Tan45°. Tan30°)
 
( 1 + √3 /3 ) /( 1 - √3/ 3) = 2 + √3
 
   Tan 15 = ( Tan 45° - Tan30° ) / (1 + √3/ 3 ) 
 
             = ( 1 - √3 / 3 )/ (1  + √3 /3 ) =
 
              =  ( 3 - √3 ) / ( 3 + √3 )   
                                    
               = 2 - √3
Hi Nick,
 
(a)   tan(A+B)=sin(A+B)/cos(A+B)
                     =sinAcosB+cosAsinB
                        cosAcosB-sinAsinB
 
divide the numerator and denominator by cosAcosB
 
                    =sinA/cosA+sinB/cosB
                         1-sinAsinB/cosAcosB
                    =tanA+tanB
                      1-tanAtanB
 
(b)   tan75=tan(30+45)
               =tan30+tan45
                 1-tan30tan45
 
               =1/√3+1
                 1-1/√3
 
               =(1+√3)/(√3-1)
 
multiply the numerator and denominator by √3+1
 
              =(1+√3)²/(√3)²-(1)²
 
              =(1+2√3+3)/(3-1)
 
              =(4+2√3)/2
              =2+√3
 
 
(c)  tan15=tan(60-45)
               =tan60-tan45
                 1+tan60tan45
 
               =(√3-1)/(1+√3)
 
multiply the numerator and denominator by 1-√3
 
  and √3-1=-(1-√3)
 
 
               =-(1-√3)²/(1)²-(√3)²
 
               =-(1-2√3+3)/-2
 
              =(4-2√3)/2
 
              =2-√3

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