Question

In triangle $ABC$, if $\cot A,\cot B,\cot C$ are in $A.P.$, then $a^2,b^2,c^2$ are in:

• A. $A.P.$
• B. $G.P.$
• C. $H.P.$
• D. $A.G.P.$

Answer 1

The correct option is A $A.P.$

Given:
$\cot A,\cot B,\cot C$ are in $A.P.$
Thus $\cot B-\cot A=\cot C-\cot B$
$\Rightarrow \frac{\sin (A-B)}{\sin A\sin B}=\frac{\sin (B-C)}{\sin B\sin C}$
$\Rightarrow \sin (A-B)\cdot \sin C=\sin (B-C)\cdot \sin A$
$\Rightarrow \sin (A-B)\cdot \sin (A+B)=\sin (B-C)\cdot \sin (B+C)$
$\Rightarrow {\sin }^{2}A-{\sin }^{2}B={\sin }^{2}B-{\sin }^{2}C$
$\Rightarrow 2{\sin }^{2}B={\sin }^{2}A+{\sin }^{2}C$
$\therefore 2b^2=a^2+c^2$
$\Rightarrow a^2,b^2,c^2$ are in $A.P.$