Question

In a $\Delta ABC,$ lf $\tan \frac{A}{2}$ , $\tan \frac{B}{2},\tan \frac{C}{2}$ are in A.P, then cos A, cos B cos C are

• A. in A.P.
• B. in G.P.
• C. in H.P.
• D. not related

Answer 1

The correct option is A in A.P.

Given

$\tan A/2,\tan B/2,\tan C/2$ are in AP

$⟹2\tan \frac{B}{2}=\tan \frac{A}{2}+\tan \frac{C}{2}$

$⟹2\left(\frac{\sin \frac{B}{2}}{\cos \frac{B}{2}}\right)=\left(\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}}\right)+\left(\frac{\sin \frac{C}{2}}{\cos \frac{C}{2}}\right)$

$⟹2\left(\frac{\sin \frac{B}{2}}{\cos \frac{B}{2}}\right)=\frac{\sin \frac{A}{2}\cos \frac{C}{2}+\cos \frac{A}{2}\sin \frac{C}{2}}{\cos \frac{A}{2}\cos \frac{C}{2}}$

$⟹2\sin \frac{B}{2}\cos \frac{A}{2}\cos \frac{C}{2}=\cos \frac{B}{2}\left(\sin \frac{A}{2}\cos \frac{C}{2}+\cos \frac{A}{2}\sin \frac{C}{2}\right)$

$(\sin (A+B)=\sin A\cos B+\cos B\sin A)$

$⟹2\sin \frac{B}{2}\cos \frac{A}{2}\cos \frac{C}{2}=\cos \frac{B}{2}\left(\sin \frac{A+C}{2}\right)$

$⟹2\sin \frac{B}{2}\cos \frac{A}{2}\cos \frac{C}{2}=\cos \frac{B}{2}\left(\sin \frac{\pi -B}{2}\right)$

$⟹2\sin \frac{B}{2}\cos \frac{A}{2}\cos \frac{C}{2}=\cos \frac{B}{2}\cos \frac{B}{2}$

$⟹4\sin \frac{B}{2}\cos \frac{A}{2}\cos \frac{C}{2}=2{\cos }^2\frac{B}{2}$

$⟹-1+4\sin \frac{B}{2}\cos \frac{A}{2}\cos \frac{C}{2}=2{\cos }^2\frac{B}{2}-1$

$(\cos 2\theta =2{\cos }^2\theta -1and\cos A+\cos C-\cos B=-1+4\sin \frac{B}{2}\cos \frac{A}{2}\cos \frac{C}{2})$

$⟹\cos A+\cos C-\cos B=\cos B$

$⟹2\cos B=\cos A+\cos B$

$⟹\cos A,\cos B,\cos C$ are in AP