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Trigonometry

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1 Answer

cotθ=k
 
tanθ=1/k
 
tan3θ=tan(θ+2θ)
        =tanθ+tan2θ
          1-tanθtan2θ
 
       =tanθ+2tanθ/1-tan²θ
          1-tanθ.2tanθ/1-tan²θ
 
        =tanθ(1-tan²θ)+2tanθ
         1-tan²θ-2tan²θ
 
        =tanθ-tan³θ+2tanθ
           1-3tan²θ
 
        =3tanθ-tan³θ
           1-3tan²θ
 
       =  3(1/k)-(1/k)³
            1-3(1/k)²
 
       =(3k²-1)/k³
         ( k²-3)/k²
 
       =(3k²-1)/(k)(k²-3)