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Grade 11Trigonometry

prove that:

tan3A.tan2A.tanA = tan3A - tan2A - tanA

Profile image of paradox xyz cool
15 Years agoGrade 11
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2 Answers

Profile image of vikas askiitian expert
ApprovedApproved Tutor Answer15 Years ago

tan2A can be written as  tan(3A-A)

 

now we have tan(a-b)=tana-tanb/1+tanatanb

here 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-tan2A

hence proved

 

Profile image of Rishi Sharma
6 Years ago
Hello student,
tan2A can be written as tan(3A-A)
now we have tan(a-b)=tana-tanb/1+tanatanb
here a =2A & b=A
tan2A=tan(3A-A)=(tan3A-tanA)/(1+tan3AtanA)
tan2A(1+tan3Atan2A) = tan3A-tanA
tan2A + tan2Atan3AtanA = tan3A -tanA
tan2Atan3AtanA = tan3A -tanA-tan2A
Hence proved
Thanks
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