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paradox xyz cool Grade: 11
        


prove


tan3A.tan2A.tanA = tan3A - tan2A - tanA


6 years ago

Answers : (2)

SAGAR SINGH - IIT DELHI
879 Points
										

Dear student,


tan(3A) = tan(2A+A)
tan(3A) = [tan(2A) + tan(A)] / [1 - tan(A)*tan(2A)]

cross multiply
tan(3A) * [1 - tan(A)*tan(2A)] = [tan(2A) + tan(A)]
tan(3A) - tan(A)*tan(2A)*tan(3A) = tan(2A) + tan(A)
tan(A)*tan(2A)*tan(3A) = tan(3A) - tan(2A) - tan(A)


 







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Askiitians Expert


Sagar Singh


B.Tech, IIT Delhi






6 years ago
vikas askiitian expert
510 Points
										

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

6 years ago
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