Given $3 tan A \cdot tan B = 1$ $3 TanA*TanB=1$ ,show that 2cos(A+B)=cos(A-B) ?

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Given $3 tan A \cdot tan B = 1$ $3 TanA*TanB=1$ ,show that 2cos(A+B)=cos(A-B) ?

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Answer 1 · 2017-06-12T06:28:35

Given

$3 tan A tan B = 1$ $3tanAtanB=1$

$\Rightarrow \frac{1}{tan A tan B} = 3$ $=>1/(tanAtanB)=3$

$\Rightarrow \frac{cos A cos B}{sin A sin B} = \frac{3}{1}$ $=>(cosAcosB)/(sinAsinB)=3/1$

By componendo and dividendo

$\Rightarrow \frac{cos A cos B + sin A sin B}{cos A cos B - sin A sin B} = \frac{3 + 1}{3 - 1}$ $=>(cosAcosB+sinAsinB)/ (cosAcosB-sinAsinB)= (3+1)/(3-1)$

$\Rightarrow \frac{cos (A - B)}{cos (A + B)} = \frac{4}{2} = 2$ $=>cos(A-B)/ cos(A+B)= 4/2=2$

$\Rightarrow 2 cos (A + B) = cos (A - B)$ $=>2cos(A+B)=cos(A-B)$

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Given $3 tan A \cdot tan B = 1$ $3 TanA*TanB=1$ ,show that 2cos(A+B)=cos(A-B) ?

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