Question

In $∆ABC$, if $cotA,cotB$, and $cotC$ are in AP, then $a^2,b^2$, and $c^2$ are in

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

Answer 1

The correct option is B

AP

Determining relationship between $a^2,b^2$, and $c^2$.

Since the numbers are in AP, the difference between the consecutive numbers is equal, therefore, we can write :

$$\begin{gathered} \Rightarrow cotB–cotA=cotC–cotB \\ \Rightarrow \frac{\cos B}{\sin B}–\frac{\cos A}{\sin A}=\frac{\cos C}{\sin C}–\frac{\cos B}{\sin B};\left[\because \cot \left(x\right)=\frac{\cos x}{\sin x}\right] \\ \Rightarrow \frac{(\sin A\cos B–\cos A\sin B)}{\sin B\sin A}=\frac{(\sin B\cos C–\cos B\sin C)}{\sin B\sin C} \\ \Rightarrow \frac{\sin (A–B)}{\sin B\sin A}=\frac{\sin (B–C)}{\sin B\sin C};\left[\because \sin (a-b)=\sin a\times \cos b-\cos a\times \sin b\right] \\ \Rightarrow \frac{\sin (A–B)}{\sin A}=\frac{\sin (B–C)}{\sin C} \\ \Rightarrow \sin (A–B)\sin C=\sin A\sin (B–C) \\ \Rightarrow \sin (A–B)\sin (A+B)=\sin (B+C)\sin (B–C);\left[\because A+B+C=180,\sin C=\sin (A+B)and\sin A=\sin (B+C)\right] \\ \Rightarrow {\sin }^{2}A–{\sin }^{2}B={\sin }^{2}B–{\sin }^{2}C \\ \Rightarrow {\sin }^{2}A+{\sin }^{2}C=2{\sin }^{2}B \\ \Rightarrow a^2+c^2=2b^2 \end{gathered}$$

Hence, option B is the correct answer.