Question

cos 2B=cos(A+C)/cos(A-C) then​

Answer 1

Your Answer !!!!!

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》If $Cos\: 2B =\:$$\frac{cos(A+C)}{cos(A-C)}$

》Then $tan \: A ,\: tan \:B,tan \: C$ are in .

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Answer 2

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$\leadsto$ $\sf cos 2B = \frac{cos(A + C)}{cos(A - C)}\:then,$

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$\leadsto$ $\sf cos 2B = \frac{cos(A + C)}{cos(A - C)}$

$\leadsto$ $\sf \frac{cos 2B}{1} = \frac{cos(A + C)}{cos(A - C)}$

$\leadsto$ $\sf \frac{cos 2B + 1}{cos 2B - 1} = \frac{cos(A + C) + cos(A - C)}{cos(A + C) - cos(A - C)}$

$\leadsto$ $\sf \frac{2 sin^{2}B}{2 cos^{2}B} = \frac{2 sin A\: sin C}{2 cos A \:cos C}$

$\leadsto$ $\sf tan^{2}B = tan A\:tan C$

$\pink\leadsto$ $\boxed{\pink{\sf \frac{tan A}{tan B} = \frac{tan B}{tan C}}}$ $\orange\star$

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