A) A.P.
B) G.P.
C) H.P.
D) None of these
Correct Answer: B
Solution :
\[\cos 2B=\frac{\cos (A+C)}{\cos (A-C)}=\frac{\cos A\cos C-\sin A\sin C}{\cos A\cos C+\sin A\sin C}\] Þ \[\frac{1-{{\tan }^{2}}B}{1+{{\tan }^{2}}B}=\frac{1-\tan A\tan C}{1+\tan A\tan C}\] Þ \[1+{{\tan }^{2}}B-\tan A\tan C-\tan A\tan C{{\tan }^{2}}B\] \[=1-{{\tan }^{2}}B+\tan A\tan C-\tan A\tan C{{\tan }^{2}}B\] Þ \[2{{\tan }^{2}}B=2\tan A\tan C\Rightarrow {{\tan }^{2}}B=\tan A\tan C\] Hence, tan A, tan B and \[\tan \]C will be in G.P.
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