In a triangle ABC,if tan(A/2),tan(B/2),and ta | vedclass

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(A) Given that in a $	riangle ABC$,$	an(A/2)$,$	an(B/2)$,and $	an(C/2)$ are in Arithmetic Progression $(AP)$.Therefore,$2 	an(B/2) = 	an(A/2) +

Vedclass · Accepted Answer

$\cos A$,$\cos B$,and $\cos C$ are in Arithmetic Progression.

Vedclass · Answer

(A) Given that in a $	riangle ABC$,$	an(A/2)$,$	an(B/2)$,and $	an(C/2)$ are in Arithmetic Progression $(AP)$.Therefore,$2 	an(B/2) = 	an(A/2) + 	an(C/2)$.Using the identity $	an(A/2) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$,this can be solved,but a more direct trigonometric approach is:$	an(B/2) - 	an(A/2) = 	an(C/2) - 	an(B/2)$$\frac{\sin(B/2)}{\cos(B/2)} - \frac{\sin(A/2)}{\cos(A/2)} = \frac{\sin(C/2)}{\cos(C/2)} - \frac{\sin(B/2)}{\cos(B/2)}$$\frac{\sin(B/2-A/2)}{\cos(B/2)\cos(A/2)} = \frac{\sin(C/2-B/2)}{\cos(C/2)\cos(B/2)}$$\frac{\sin((B-A)/2)}{\cos(A/2)} = \frac{\sin((C-B)/2)}{\cos(C/2)}$Using $A/2 = 90^{\circ} - (B+C)/2$ and $C/2 = 90^{\circ} - (A+B)/2$,we get:$\frac{\sin((B-A)/2)}{\sin((B+C)/2)} = \frac{\sin((C-B)/2)}{\sin((A+B)/2)}$$\sin((A+B)/2)\sin((B-A)/2) = \sin((B+C)/2)\sin((C-B)/2)$Using $2\sin x \sin y = \cos(x-y) - \cos(x+y)$:$\cos A - \cos B = \cos B - \cos C$$\cos A + \cos C = 2\cos B$Thus,$\cos A$,$\cos B$,and $\cos C$ are in Arithmetic Progression.

In a $\triangle ABC$,if $\tan(A/2)$,$\tan(B/2)$,and $\tan(C/2)$ are in Arithmetic Progression,then which of the following options is always correct?

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