In a $\triangle A B C$, if $\tan (A / 2), \tan (B / 2)$ and $\tan (C / 2)$ are in Arithmetic progression,…
In a $\triangle A B C$, if $\tan (A / 2), \tan (B / 2)$ and $\tan (C / 2)$ are in Arithmetic progression, then which of the following option is always correct?
cos A, cosB and cosC are in Arithmetic Progression.
cos A, cosB and cosC are in Geometric Progression.
cos A, cosB and cosC are in harmonic Progression.
no conclusion can be made with given data.
Solution
Given In a $\triangle A B C, \tan (A / 2), \tan (B / 2), \tan (C / 2)$ are in Arithmetic progression,
$\begin{aligned} & \therefore \tan (B / 2)-\tan (A / 2)=\tan (C / 2)-\tan (B / 2) \\ & \frac{\sin (B / 2)}{\cos (B / 2)}-\frac{\sin (A / 2)}{\cos (A / 2)} \\ & \quad=\frac{\sin (C / 2)}{\cos (C / 2)}-\frac{\sin (B / 2)}{\cos (B / 2)} \\ & \frac{\sin (B / 2) \cdot \cos (A / 2)-\sin (A / 2) \cdot \cos (B / 2)}{\cos (B / 2) \cdot \cos (A / 2)} \\ & \quad=\frac{\sin (C / 2) \cdot \cos (B / 2)-\sin (B / 2) \cos (C / 2)}{\cos (C / 2) \cdot \cos (B / 2)} \\ & \quad \frac{\sin \left(\frac{B}{2}-\frac{A}{2}\right)}{\cos (A / 2)}=\frac{\sin \left(\frac{C}{2}-\frac{B}{2}\right)}{\cos (C / 2)}...(i)\end{aligned}$.
$\because A+B+C=180^{\circ}$ (angle sum property in $\triangle A B C$ )
$
\begin{gathered}
\frac{A}{2}=\frac{1}{2}\{180-(B+C)\} \\
\frac{A}{2}=90-\left(\frac{B}{2}+\frac{C}{2}\right) \text { and } \frac{C}{2}=90-\left(\frac{A}{2}+\frac{B}{2}\right)
\end{gathered}
$
From Eq. (i), we get
$
\begin{aligned}
& \frac{\sin \left(\frac{B}{2}-\frac{A}{2}\right)}{\cos \left\{90-\left(\frac{B}{2}+\frac{C}{2}\right)\right\}}=\frac{\sin \left(\frac{C}{2}-\frac{B}{2}\right)}{\cos \left\{90-\left(\frac{A}{2}+\frac{B}{2}\right)\right\}} \\
& \Rightarrow \frac{\sin \left(\frac{B}{2}-\frac{A}{2}\right)}{\sin \left(\frac{B}{2}+\frac{C}{2}\right)}=\frac{\sin \left(\frac{C}{2}-\frac{B}{2}\right)}{\sin \left(\frac{B}{2}+\frac{A}{2}\right)} \\
& \Rightarrow \sin \left(\frac{B}{2}+\frac{A}{2}\right) \cdot \sin \left(\frac{B}{2}-\frac{A}{2}\right) \\
& =\sin \left(\frac{C}{2}+\frac{B}{2}\right) \cdot \sin \left(\frac{C}{2}-\frac{B}{2}\right) \\
& \Rightarrow 2 \sin \left(\frac{B}{2}+\frac{A}{2}\right) \cdot \sin \left(\frac{B}{2}-\frac{A}{2}\right) \\
& =2 \sin \left(\frac{C}{2}+\frac{B}{2}\right) \cdot \sin \left(\frac{C}{2}-\frac{B}{2}\right) \\
& \{\because 2 \sin x \cdot \sin y=\cos (x-y)-\cos (x+y)\} \\
& \Rightarrow \cos \left(\frac{B}{2}+\frac{A}{2}-\frac{B}{2}+\frac{A}{2}\right)-\cos \left(\frac{B}{2}+\frac{A}{2}+\frac{B}{2}-\frac{A}{2}\right) \\
& =\cos \left(\frac{C}{2}+\frac{B}{2}-\frac{C}{2}+\frac{B}{2}\right)-\cos \left(\frac{C}{2}+\frac{B}{2}+\frac{C}{2}-\frac{B}{2}\right) \\
& \cos A-\cos B=\cos B-\cos C \\
& \Rightarrow 2 \cos B=\cos A+\cos C \\
& \Rightarrow \cos A, \cos B \text { and } \cos C \text { are in AP. } \\
&
\end{aligned}
$