Question

In any triangle $ABC,\sin A,\sin B,\sin C$ are in A.P., then the maximum value of $\tan \frac{B}{2}$ is

• A. $\frac{1}{3}$
• B. $\frac{1}{\sqrt{3}}$
• C. $-\frac{1}{\sqrt{3}}$
• D. none of these

Answer 1

Answer: (B)

$\frac{1}{\sqrt{3}}$

We have, $\sin A,\sin B,\sin C$ are in $A.P.$

$\Rightarrow 2\sin B=\sin A+\sin C=\sin A+\sin (A+B)$$(\because A+B+C={180}^0\Rightarrow C={180}^0-(A+B))$

$=\sin A+\sin A\cos B+\cos A\sin B$

$\Rightarrow \sin B(2-\cos A)=\sin A(1+\cos B)$

$\Rightarrow \frac{\sin B}{1+\cos B}=\frac{\sin A}{2-\cos A}$

$\Rightarrow \tan \frac{B}{2}=\frac{\sin A}{1-\cos A}=k$ (say)

$\Rightarrow \sin A=2k-k\cos A\Rightarrow \sin A+k\cos A=2k$

$\Rightarrow \sin (A+\alpha )=\frac{2k}{\sqrt{1+k^2}},$ where $\tan \alpha =\frac{k}{1}$

$\Rightarrow \frac{2k}{\sqrt{1+k^2}}\le 1\Rightarrow 2k\le \sqrt{1+k^2}\Rightarrow 4k^2\le 1+k^2$

$\Rightarrow 3k^2-1\le 0\Rightarrow \left|k\right|\le \frac{1}{\sqrt{13}}$

$\Rightarrow$ Maximum value of $\tan \frac{B}{2}$ is $\frac{1}{\sqrt{3}}.$