Question

If $a,b,c$ are the sides and $A,B,C$ are the angles of a triangle $ABC$, then $\tan \dfrac{A}{2}$ is equal to
A. $\sqrt{\dfrac{(s-c)(s-a)}{s(s-b)}}$
B. $\sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}$
C. $\sqrt{\dfrac{(s-a)(s-b)}{s(s-c)}}$
D. $\sqrt{\dfrac{(s-a)s}{(s-b)(s-c)}}$

Answer 1

Hint: To solve this question, we will use the formula of half angle of tan in terms of the semi perimeter of the triangle.
We will use the formula of semi perimeter of triangle, cosine rule for the angle $A$ that is$\cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}$and trigonometric formulas. We will calculate the value of the half angles for sine and cosine and then divide them to get the value of $\tan \dfrac{A}{2}$as we know$\tan \dfrac{A}{2}=\dfrac{\sin \dfrac{A}{2}}{\cos \dfrac{A}{2}}$.

Formula Used: The half angle formulas are:
$\begin{align}
  & {{\sin }^{2}}\dfrac{A}{2}=\dfrac{1-\cos A}{2} \\
 & {{\cos }^{2}}\dfrac{A}{2}=\dfrac{1+\cos A}{2} \\
\end{align}$
The semi perimeter of the triangle is:
$\begin{align}
  & s=\dfrac{a+b+c}{2} \\
 & 2s=a+b+c \\
\end{align}$

Complete step by step solution: We are given a triangle $ABC$ having sides $a,b,c$ and angles $A,B,C$ and we have to determine the value of $\tan \dfrac{A}{2}$.
We will use the formula of tan $\tan \dfrac{A}{2}=\dfrac{\sin \dfrac{A}{2}}{\cos \dfrac{A}{2}}$ to calculate its value.
First we will calculate the value of $\sin \dfrac{A}{2}$ and $\cos \dfrac{A}{2}$.
To calculate the value of $\sin \dfrac{A}{2}$, we will take ${{\sin }^{2}}\dfrac{A}{2}$. Now we can write ${{\sin }^{2}}\dfrac{A}{2}$ as,
  ${{\sin }^{2}}\dfrac{A}{2}=\dfrac{1-\cos A}{2}$
Using cosine rule we will substitute the value $\cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}$.
\[\begin{align}
  & {{\sin }^{2}}\dfrac{A}{2}=\dfrac{1-\cos A}{2} \\
 & =\dfrac{1-\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}}{2} \\
 & =\dfrac{{{a}^{2}}-{{b}^{2}}-{{c}^{2}}+2bc}{4bc} \\
 & =\dfrac{{{a}^{2}}-{{(b-c)}^{2}}}{4bc}
\end{align}\]
We will now use the formula ${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$,
\[\begin{align}
  & {{\sin }^{2}}\dfrac{A}{2}=\dfrac{(a-(b-c))(a+b-c)}{4bc} \\
 & =\dfrac{(a-b+c)(a+b-c)}{4bc} \\
 & =\dfrac{(a+b+c-2b)(a+b+c-2c)}{4bc}
\end{align}\]
We will now use the formula of the semi perimeter of the triangle and substitute it in the above equation.
\[\begin{align}
  & {{\sin }^{2}}\dfrac{A}{2}=\dfrac{(2s-2b)(2s-2c)}{4bc} \\
 & {{\sin }^{2}}\dfrac{A}{2}=\dfrac{4(s-b)(s-c)}{4bc} \\
 & \sin \dfrac{A}{2}=\sqrt{\dfrac{(s-b)(s-c)}{bc}}
\end{align}\]
We will now find the value of $\cos \dfrac{A}{2}$ . We will take ${{\cos }^{2}}\dfrac{A}{2}$ and write it as ${{\cos }^{2}}\dfrac{A}{2}=\dfrac{1+\cos A}{2}$ and substitute the value $\cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}$ .
\[\begin{align}
  & {{\cos }^{2}}\dfrac{A}{2}=\dfrac{1+\cos A}{2} \\
 & =\dfrac{1+\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}}{2} \\
 & =\dfrac{2bc+{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{4bc} \\
 & =\dfrac{{{(b+c)}^{2}}-{{a}^{2}}}{4bc} \\
 & =\dfrac{(b+c+a)(b+c-a)}{4bc} \\
 & =\dfrac{(b+c+a)(b+c+a-2a)}{4bc}
\end{align}\]
We will now substitute the formula of the semi perimeter of the triangle in the equation.
\[\begin{align}
  & {{\cos }^{2}}\dfrac{A}{2}=\dfrac{2s(2s-2a)}{4bc} \\
 & \cos \dfrac{A}{2}=\sqrt{\dfrac{4s(s-a)}{4bc}} \\
 & \cos \dfrac{A}{2}=\dfrac{\sqrt{s(s-a)}}{bc}
\end{align}\]
We will now find the value of $\tan \dfrac{A}{2}$.
$\begin{align}
  & \tan \dfrac{A}{2}=\dfrac{\sin \dfrac{A}{2}}{\cos \dfrac{A}{2}} \\
 & =\dfrac{\sqrt{\dfrac{(s-b)(s-c)}{bc}}}{\sqrt{\dfrac{s(s-a)}{bc}}} \\
 & =\sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}
\end{align}$

The value of $\tan \dfrac{A}{2}$ for the triangle $ABC$ is $\tan \dfrac{A}{2}=\sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}$ when $a,b,c$ are the sides and $A,B,C$ are the angles. Hence the correct option is (B).


Note: This $\tan \dfrac{A}{2}=\sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}$is the formula of half angle of tan and we could have directly chosen it from the options.