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```
prove
tan3A.tan2A.tanA = tan3A - tan2A - tanA
```
7 years ago

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)

All the best.
Win exciting gifts by      answering the questions on Discussion Forum. So help discuss any    query   on askiitians forum and become an Elite Expert League askiitian.

Sagar Singh
B.Tech, IIT Delhi

```
7 years ago
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
```
7 years ago
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