How to Prove tan(a - b) = tana - tanb without using a prefix?

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Homework Help Overview

The discussion revolves around proving the trigonometric identity tan(a - b) = (tan a - tan b) / (1 + tan a tan b) without using a prefix. The subject area includes trigonometric identities and their derivations.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using known trigonometric identities for sine and cosine to manipulate the expression for tan(a - b). There is an attempt to simplify the expression by plugging in these identities, but some participants express uncertainty about the next steps. Questions arise about how to effectively introduce a factor to achieve the desired form.

Discussion Status

Some guidance has been offered regarding the manipulation of the numerator and denominator to achieve the necessary form for the proof. Participants are exploring different approaches to simplify the expression, with a focus on dividing by cos(a)cos(b) to facilitate the proof.

Contextual Notes

There is an emphasis on not using a prefix in the proof, which may limit the methods available for participants to explore. The original poster expresses being stumped at a certain point in the derivation.

Quinn Morris
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Homework Statement



Prove that: tan (a - b) = tana - tanb
1 + tanatanb
]​

Homework Equations



cos (a- b) = cosacosb + sinasinb


sin (a - b) = sinacosb - cosasinb


The Attempt at a Solution



I have plugged in the two equations give above since tan - sin/cos and then I'm am stumped of how to proceed

tan (a - b) = sinacosb - cosasinb
cosacosb + sinasinb

can anyone help please?
 
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Quinn Morris said:
tan (a - b) = sinacosb - cosasinb
cosacosb + sinasinb

The formula you are trying to arrive at provides the clue: what could you divide numerator and denominator by to get that "1" in the denominator?
 
so times buy a form of one... 1 over cosacosb?
 
yes that is correct, divide the numerator and denominator by cosacosb
 

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