Question

In any triangle ABC, $\sin A,\sin B,\sin C$ are in A.P. Find the maximum value of $\tan \dfrac{B}{2}$.
A. $\dfrac{1}{3}$
B. $\dfrac{1}{\sqrt{3}}$
C. $\dfrac{-1}{\sqrt{3}}$
D. none of these

Answer 1

We need to first try to find the value of $\tan \dfrac{B}{2}$ using different trigonometric formulas. We find the value with respect to sin. We try to find the range of sin value. Using the range and a variable we find the maximum value of $\tan \dfrac{B}{2}$.

Complete step by step answer:
We know that if a, b, c is in A.P. then we can say that $a+c=2b$.
It’s given that for triangle ABC, $\sin A,\sin B,\sin C$ are in A.P. which tells us $\sin A+\sin C=2\sin B........(i)$.
We have the trigonometric formula that $\sin A+\sin C=2\sin \left( \dfrac{A+C}{2} \right)\cos \left( \dfrac{A-C}{2} \right)$.
We know that A, B, C are the angles of $\Delta ABC$.
From the angle law of triangles, we can say that $A+B+C=\pi$.
This gives us $C=\pi -\left( A+B \right)$. We also know that $\sin B=2\sin \dfrac{B}{2}\cos \dfrac{B}{2}$.
So, $\sin \left[ \pi -\left( A+B \right) \right]=\sin \left[ 2\times \dfrac{\pi }{2}-\left( A+B \right) \right]=\sin \left( A+B \right)$
Putting these values, we get $\sin A+\sin C=2\sin B........(i)$

& \sin A+\sin C=2\sin B \\\ & \Rightarrow \sin A+\sin \left[ \pi -\left( A+B \right) \right]=2\sin B \\\ & \Rightarrow \sin A+\sin \left( A+B \right)=2\sin B \\\ \end{aligned}$$ We know that $$\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$$. We have $$\left( 1+\cos B \right)=2{{\cos }^{2}}\dfrac{B}{2}$$. We replace the value and get $$\begin{aligned} & \sin A+\sin \left( A+B \right)=2\sin B \\\ & \Rightarrow \sin A+\sin A\cos B+\cos A\sin B=2\sin B \\\ & \Rightarrow \sin B\left( 2-\cos A \right)=\sin A\left( 1+\cos B \right) \\\ \end{aligned}$$ Now we take similar angles on same sides $$\begin{aligned} & \sin B\left( 2-\cos A \right)=\sin A\left( 1+\cos B \right) \\\ & \Rightarrow \dfrac{\sin B}{\left( 1+\cos B \right)}=\dfrac{\sin A}{\left( 2-\cos A \right)} \\\ & \Rightarrow \dfrac{2\sin \dfrac{B}{2}\cos \dfrac{B}{2}}{2{{\cos }^{2}}\dfrac{B}{2}}=\dfrac{\sin A}{\left( 2-\cos A \right)} \\\ & \Rightarrow \tan \dfrac{B}{2}=\dfrac{\sin A}{\left( 2-\cos A \right)} \\\ \end{aligned}$$ Let’s assume $$\dfrac{\sin A}{\left( 2-\cos A \right)}=k$$. So, we get $$\begin{aligned} & \dfrac{\sin A}{\left( 2-\cos A \right)}=k \\\ & \Rightarrow \sin A=k\left( 2-\cos A \right) \\\ & \Rightarrow \sin A+k\cos A=2k \\\ \end{aligned}$$ Now we convert the left into one single trigonometric function. We divide both sides with $\sqrt{{{k}^{2}}+1}$. $$\begin{aligned} & \sin A+k\cos A=2k \\\ & \Rightarrow \dfrac{1}{\sqrt{{{k}^{2}}+1}}\sin A+\dfrac{k}{\sqrt{{{k}^{2}}+1}}\cos A=\dfrac{2k}{\sqrt{{{k}^{2}}+1}} \\\ \end{aligned}$$ Now if we consider $$\dfrac{1}{\sqrt{{{k}^{2}}+1}}=\cos \alpha $$, then we get $$\dfrac{k}{\sqrt{{{k}^{2}}+1}}=\sin \alpha $$. $$\begin{aligned} & \dfrac{1}{\sqrt{{{k}^{2}}+1}}\sin A+\dfrac{k}{\sqrt{{{k}^{2}}+1}}\cos A=\dfrac{2k}{\sqrt{{{k}^{2}}+1}} \\\ & \Rightarrow \sin \left( A+\alpha \right)=\dfrac{2k}{\sqrt{{{k}^{2}}+1}} \\\ \end{aligned}$$ We know that for any value of x, $-1\le \sin x\le 1$. So, $-1\le \dfrac{2k}{\sqrt{{{k}^{2}}+1}}\le 1$. We need to find the maximum value of $\tan \dfrac{B}{2}$ which is equal to k. $$\begin{aligned} & \dfrac{2k}{\sqrt{{{k}^{2}}+1}}\le 1 \\\ & \Rightarrow 4{{k}^{2}}\le {{k}^{2}}+1 \\\ & \Rightarrow \left| k \right|\le \dfrac{1}{\sqrt{3}} \\\ \end{aligned}$$ **So, the correct answer is “Option B”.** **Note:** We need to remember that the way we found the value of $\sqrt{{{k}^{2}}+1}$ is the root of the sum of the squares of the coefficients. In any case if the form is like that, we can convert that to both cos and sin in values. We don’t need to consider the case $-1\le \dfrac{2k}{\sqrt{{{k}^{2}}+1}}$ as we are taking the modulus value.