In a △ABC,tan⁡A2=56,tan⁡C2=25,then

A
$a, c, b$ are in A.P.
B
$a, b, c$ are in A.P.
C
$b, a, c$ are in AP.
D
$a, b, c$ are in G.P.

Solution:

We have,

$\begin{array}{l} \tan \frac{A}{2} = \frac{5}{6} and, \tan \frac{C}{2} = \frac{2}{5} \\ \Rightarrow \tan \frac{A}{2} \tan \frac{C}{2} = \frac{1}{3} \\ \Rightarrow \sqrt{\frac{(s - b) (s - c)}{s (s - a)} \cdot \frac{(s - b) (s - a)}{s (s - c)}} = \frac{1}{3} \\ \Rightarrow \frac{s - b}{s} = \frac{1}{3} \\ \Rightarrow 3 s - 3 b = s \end{array}$

$\Rightarrow 2 s = 3 b \Rightarrow 2 b = a + c \Rightarrow a, b, c$ are in A.P.

In a $△ A B C, \tan \frac{A}{2} = \frac{5}{6}, \tan \frac{C}{2} = \frac{2}{5},$ then

Solution:

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