Question

In a triangle if $\tan \left(A/2\right),\tan \left(B/2\right),\tan \left(C/2\right)$ are in A.P., then $\cos A,\cos B,\cos C$ are also in A.P.

Answer 1

$\tan \left(A/2\right)-\tan \left(B/2\right)=\tan \left(B/2\right)-\tan \left(C/2\right)$
or $\frac{\sin \left\{\right(A-B\left)/2\right\}}{\cos \left(A/2\right)\cos \left(B/2\right)}=\frac{\sin \left\{\right(B-C\left)/2\right\}}{\cos \left(B/2\right)\cos \left(C/2\right)}$
or $\sin \left\{\right(A-B\left)/2\right\}\cdot \cos \left(C/2\right)=\sin \left\{\right(B-C\left)/2\right\}\cos \left(A/2\right)$
or $2\sin \left(\frac{A-B}{2}\right)\sin \left(\frac{A+B}{2}\right)=2\sin \left(\frac{B-C}{2}\right)\sin \left(\frac{B+C}{2}\right)$
or $\cos B-\cos A=\cos C-\cos B$
$\therefore \cos A,\cos B,\cos C$ are in A.P.