A) A.P.
B) G.P.
C) H.P.
D) None
Correct Answer: C
Solution :
\[\tan A/2,\tan B/2,\tan C/2\]in A.P. \[\Rightarrow \]\[\tan A/2,\tan B/2=\tan B/2-\tan C/2\] \[\Rightarrow \]\[\frac{\sin A/2}{\cos A/2}-\frac{\sin B/2}{\cos B/2}=\frac{\sin B/2}{\cos B/2}-\frac{\sin C/2}{\cos C/2}\] \[\Rightarrow \]\[\cos (C/2)\sin (A/2-B/2)\] \[=\cos A/2\sin (B/2-C/2)\] \[\Rightarrow \]\[\cos B-\cos A=\cos C-\cos B\] \[\Rightarrow \]\[2\cos B=\cos A+\cos C\] \[\Rightarrow \]\[\sec A,\sec B,\sec C\]are in A.P.
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