Question

In any $△ABC$, if $\cot \frac{A}{2},\cot \frac{B}{2},\cot \frac{C}{2}$ are in A.P, then $a^2,b^2,c^2$ are in (Here, $A,B,C$ and $a,b,c$ have their ususal meanings.)

• A. AP
• B. GP
• C. HP
• D. None of these

Answer 1

The correct option is A AP

Here,

$\cot \frac{A}{2},\cot \frac{B}{2},\cot \frac{C}{2}$ are in A.P..

So,

$2\cot \frac{B}{2}=\cot \frac{A}{2}+\cot \frac{C}{2}$

Now,

$2\frac{\cos \frac{B}{2}}{\sin \frac{B}{2}}$=$\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}$+$\frac{\cos \frac{C}{2}}{\sin \frac{C}{2}}$ ..........(1)

We know that

$\frac{\sin \frac{A}{2}}{a}=$$\frac{\sin \frac{B}{2}}{b}=$$\frac{\sin \frac{C}{2}}{c}=k$ (say)

$\sin \frac{A}{2}=ka$, $\sin \frac{B}{2}=kb,$ $\sin \frac{C}{2}=kc$

Therefore, from the equation (1), we get

$\frac{2}{bk}\times \frac{a^2+c^2-b^2}{2ac}$ $=\frac{1}{ak}\times \frac{b^2+c^2-a^2}{2bc}$+$\frac{1}{ck}\times \frac{a^2+b^2-c^2}{2ab}$

$2(a^2+c^2-b^2)=(b^2+c^2-a^2)+(a^2+b^2-c^2)$

$a^2+c^2=2b^2$

Therefore, $a^2,b^2,c^2$ are in A.P.