Question

Prove tanA.tan2A.tan4A= tan3A

Answer 1

Solution:

we may write tan2A  as  tan(3A-A)

 we know that

tan(A-B) = tanA - tanB/1+tanA Χ tanbB

Here  A =2A & B =A

Therefore  

tan2A = tan(3A-A) = tan3A-tanA/1+tan3AtanA

 

tan2A(1+tan3A Χ tan2A) = tan3A-tanA

 or.   tan2A + tan2A Χtan3A Χ tanA = tan3A -tanA    

  or, tan2A Χtan3A Χ tanA = tan3A -tanA-tan2A  =tan 3A

hence proved

Answer 2

Step-by-step explanation:

can be written as tan(3A-A) now we have tan(a-b)=tana-tanb/1+tanatanbhere a =2A & b=A tan2A=tan(3A-A)=tan3A-tanA/1+tan3AtanA tan2A(1+tan3Atan2A) = tan3A-tanA tan2A + tan2Atan3AtanA = tan3A -tanA or tan2Atan3AtanA = tan3A -tanA-tan2Ahence proved