Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $a^2$, $b^2$, $c^2$ are in A.P., then $\tan A$, $\tan B$, $\tan C$ are in?

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

Answer 1

The correct option is B H.P.

Let $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k$
Gives $\sin A=ak,\sin B=bk,\sin C=ck$ ...(1)
Now as $a^2,b^2,c^2$ is in A.P
$2b^2=a^2+c^2\Rightarrow b^2=a^2+c^2-b^2\frac{b^2}{2ac}=\frac{a^2+c^2-b^2}{2ac}\Rightarrow \frac{b}{2ac}=\frac{a^2+c^2-b^2}{2abc}\Rightarrow \frac{a\cos C+c\cos A}{2ac}=\frac{\cos B}{b}\Rightarrow \frac{2\cos B}{b}=\frac{\cos C}{c}+\frac{\cos A}{a}$
Dividing both sides by k
$\frac{2\cos B}{bk}=\frac{\cos C}{ck}+\frac{\cos A}{ak}$
Substituting value from (1)
$\frac{2\cos B}{\sin B}=\frac{\cos C}{\sin C}+\frac{\cos A}{\sin A}\Rightarrow 2\cot B=\cot C+\cot A$
$\Rightarrow \cot A,\cot B,\cot C$ is in A.P
Therefore, $\tan A,\tan B,\tan C$ is in H.P