Prove that $\frac{sin (A + B)}{cos (A - B)} = \frac{tan A - tan B}{1 + tan A tan B}$ $(sin(A+B))/(cos(A-B)) = (tan A-tan B)/(1+tan A tan B)$ ?

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Prove that $\frac{sin (A + B)}{cos (A - B)} = \frac{tan A - tan B}{1 + tan A tan B}$ $(sin(A+B))/(cos(A-B)) = (tan A-tan B)/(1+tan A tan B)$ ?

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Answer 1 · 2018-04-12T16:30:53

$\frac{tan A - tan B}{1 + tan A tan B} = tan (A - B) = \frac{sin (A - B)}{cos (A - B)}$ $(tanA-tanB)/(1+tanAtanB)=tan(A-B)=sin(A-B)/cos(A-B)$ .

Check the Problem.

Answer 2 · 2018-04-12T16:30:55

Please, see the proof below

Explanation:

$Reminder$ $"Reminder"$

$sin (A + B) = sin A cos B + sin B cos A$ $sin(A+B)=sinAcosB+sinBcosA$

$cos (A - B) = cos A cos B + sin A sin B$ $cos(A-B)=cosAcosB+sinAsinB$

Therefore,

$L H S = \frac{sin (A + B)}{cos (A - B)}$ $LHS=sin(A+B)/cos(A-B)$

$= \frac{sin A cos B + sin B cos A}{cos A cos B + sin A sin B}$ $=(sinAcosB+sinBcosA)/(cosAcosB+sinAsinB)$

$= \frac{\frac{sin A cos B}{cos A cos B} + \frac{sin B cos A}{cos A cos B}}{\frac{cos A cos B}{cos A cos B} + \frac{sin A sin B}{cos a cos B}}$ $=((sinAcosB)/(cosAcosB)+(sinBcosA)/(cosAcosB))/((cosAcosB)/(cosAcosB)+(sinAsinB)/(cosacosB))$

$= \frac{tan A + tan B}{1 + tan A tan B}$ $=(tanA+tanB)/(1+tanAtanB)$

$= R H S ?$ $=RHS?$

$Q E D$ $QED$

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Prove that $\frac{sin (A + B)}{cos (A - B)} = \frac{tan A - tan B}{1 + tan A tan B}$ $(sin(A+B))/(cos(A-B)) = (tan A-tan B)/(1+tan A tan B)$ ?

2 Answers

Explanation:

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