Question

If $a^2,b^2,c^2$are in AP, then which of the following is also an AP $?$

• A. $\sin A,\sin B,\sin C$
• B. $\tan A,\tan B,\tan C$
• C. $cotA,cotB,cotC$
• D. None of these

Answer 1

The correct option is C

$cotA,cotB,cotC$

Finding the series in AP:

Step 1. Find the relation using common difference in terms of AP

Given $a^2,b^2,c^2$are in AP, So

$b^2-a^2=c^2-b^2...\left(i\right)$

Now, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$ (sine rule)

Thus, $a=k\sin A,b=k\sin B$and $c=k\sin C$

Substitute $a,b,c$in $\left(i\right)$

Step2. Simplify the terms using identities

$\Rightarrow$ $k^2{\sin }^{2}B-k^2{\sin }^{2}A=k^2{\sin }^{2}C-k^2{\sin }^{2}B$

$\Rightarrow$ ${\sin }^{2}B-{\sin }^{2}A={\sin }^{2}C-{\sin }^{2}B$ $($ since ${\sin }^{2}A-{\sin }^{2}B=\sin (A+B)\sin (A-B))$

$\Rightarrow$ $\sin (B+A)\sin (B-A)=\sin (C+B)\sin (C-B)$

$\Rightarrow$ $\sin (\mathrm{\pi }-\mathrm{C})\sin (B-A)=\sin (\mathrm{\pi }-\mathrm{A})\sin (C-B)$ $($ since $A+B+C=\mathrm{\pi })$

$\Rightarrow$ $\sin C\sin (B-A)=\sin A\sin (C-B)$

$\Rightarrow$ $\frac{\sin (B-A)}{\sin A}=\frac{\sin (C-B)}{\sin C}$

$\Rightarrow$ $\frac{\sin (B-A)}{\sin A\sin B}=\frac{\sin (C-B)}{\sin B\sin C}$

$$\begin{gathered} \Rightarrow \frac{\sin B\cos A-\cos B\sin A}{\sin A\sin B}=\frac{\sin C\cos B-\cos C\sin B}{\sin B\sin C} \\ \Rightarrow cotA-cotB=cotB-\mathrm{co}tC \end{gathered}$$

Thus$cotA,cotB,cotC$ are in AP.

Hence, the correct option is (C).