Prove/verify the identity? $\frac{sin (α - β)}{cos (α + β)} = \frac{tan α - tan β}{1 - tan α tan β}$ $sin(alpha-beta)/(cos(alpha+beta)) = (tanalpha-tanbeta)/(1-tanalphatanbeta)$

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Prove/verify the identity? $\frac{sin (α - β)}{cos (α + β)} = \frac{tan α - tan β}{1 - tan α tan β}$ $sin(alpha-beta)/(cos(alpha+beta)) = (tanalpha-tanbeta)/(1-tanalphatanbeta)$

Thanks in advance

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Answer 1 · 2018-04-07T23:34:55

verified below...

Explanation:

$\frac{sin (a - b)}{cos (a + b)} = \frac{tan a - tan b}{1 - tan a tan b}$ $sin(a-b)/cos(a+b)= (tana-tanb)/(1-tanatanb)$

Apply sine difference and cosine sum identities:
$\frac{sin a cos b - cos a sin b}{cos a cos b - sin a sin b} = \frac{tan a - tan b}{1 - tan a tan b}$ $(sinacosb-cosasinb)/(cosacosb-sinasinb)= (tana-tanb)/(1-tanatanb)$

Divide each term in the numerator by $cos a cos b$ $cosacosb$ :
$(cosacosb((sinacancel(cosb))/(cosacancel(cosb))-(cancel(cosa)sinb)/(cancel(cosa)cosb)))/(cosacosb-sinasinb)= (tana-tanb)/(1-tanatanb)$

Apply quotient identity: $sintheta/costheta=tantheta$
$(cosacosb(tana-tanb))/(cosacosb-sinasinb)= (tana-tanb)/(1-tanatanb)$

Divide each term in the denominator by $cosacosb$ :
$(cosacosb(tana-tanb))/(cosacosb((cancel(cosacosb))/(cancel(cosacosb))-(sinasinb)/(cosacosb)))= (tana-tanb)/(1-tanatanb)$

Apply quotient identity: $sintheta/costheta=tantheta$ :
$(cancel(cosacosb)(tana-tanb))/(cancel(cosacosb)(1-tanatanb))= (tana-tanb)/(1-tanatanb)$

$(tana-tanb)/(1-tanatanb)= (tana-tanb)/(1-tanatanb)$

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Prove/verify the identity? $\frac{sin (α - β)}{cos (α + β)} = \frac{tan α - tan β}{1 - tan α tan β}$ $sin(alpha-beta)/(cos(alpha+beta)) = (tanalpha-tanbeta)/(1-tanalphatanbeta)$

Thanks in advance

1 Answer

Explanation:

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Prove/verify the identity? sin(alpha-beta)/(cos(alpha+beta)) = (tanalpha-tanbeta)/(1-tanalphatanbeta)sin(α−β)cos(α+β)=tanα−tanβ1−tanαtanβsin(alpha-beta)/(cos(alpha+beta)) = (tanalpha-tanbeta)/(1-tanalphatanbeta)

Thanks in advance

1 Answer

Explanation:

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Prove/verify the identity? $\frac{sin (α - β)}{cos (α + β)} = \frac{tan α - tan β}{1 - tan α tan β}$ $sin(alpha-beta)/(cos(alpha+beta)) = (tanalpha-tanbeta)/(1-tanalphatanbeta)$