cos2B=cos(A+C)cos(A−C) then tanA,tanB,tanC are in

1−tan2B1+tan2B=cos(A+C)cos(A−C)⇒1+tan2B1−tan2B=cos(A−C)cos(A+C)Apply C and D 1+tan2B-1+tan2B1+tan2B+1-tan2B=cos(A-B)-cos(A+B)cos(A-B)+cos(A+B) 2tan2B2=2sinAsinC2cosAcosC tan2B=tanAtanC

cos2B=cos(A+C)cos(A−C) then tanA,tanB,tanC are in

1−tan2B1+tan2B=cos(A+C)cos(A−C)⇒1+tan2B1−tan2B=cos(A−C)cos(A+C)Apply C and D 1+tan2B-1+tan2B1+tan2B+1-tan2B=cos(A-B)-cos(A+B)cos(A-B)+cos(A+B) 2tan2B2=2sinAsinC2cosAcosC tan2B=tanAtanC

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$\cos 2 B = \frac{\cos (A + C)}{\cos (A - C)}$ then $\tan A, \tan B, \tan C$ are in

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Detailed Solution

$\frac{1 - \tan^{2} B}{1 + \tan^{2} B} = \frac{\cos (A + C)}{\cos (A - C)} \Rightarrow \frac{1 + \tan^{2} B}{1 - \tan^{2} B} = \frac{\cos (A - C)}{\cos (A + C)}$

Apply C and D
$\frac{1 + \tan^{2} B - 1 + \tan^{2} B}{1 + \tan^{2} B + 1 - \tan^{2} B} = \frac{\cos (A - B) - \cos (A + B)}{\cos (A - B) + \cos (A + B)} \frac{2 \tan^{2} B}{2} = \frac{2 \sin A \sin C}{2 \cos A \cos C} \tan^{2} B = \tan A \tan C$