Question

In a triangle $\vartriangle ABC$_{, if $2s=a+b+c$, then the value of $\frac{s(s-a)}{bc}-\frac{(s-c)(s-a)}{bc}=$}
A. \[\sin A\]
B. $\cos A$
C. $\tan A$
D. None of these.

Answer 1

Hint: To solve this question we will derive the value of from equation $2s=a+b+c$_{and then put it} into the equation $\frac{s(s-a)}{bc}-\frac{(s-c)(s-a)}{bc}$_{. We will then simplify the equation using} expansion formulas and other arithmetical operations and get an equation which will be formula for some angle.
Formula Used:
The expansion formulas are:
$\begin{align}
  & {{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\
 & {{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\
 & (a+b)(a-b)={{a}^{2}}-{{b}^{2}}
\end{align}$
The cosine rule or Law of cosine are:
\[\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}=\cos A\]
Complete step-by-step solution:
We are given a triangle$\vartriangle ABC$_and we have to find the value of equation $\frac{s(s-a)}{bc}-\frac{(s-c)(s-a)}{bc}$ when $2s=a+b+c$_.
We will first determine the value of $s$,
$s=\frac{a+b+c}{2}$
Now we will substitute this value of $s$in the given equation $\frac{s(s-a)}{bc}-\frac{(s-c)(s-a)}{bc}$_{,
$\frac{\frac{a+b+c}{2}\left( \frac{a+b+c}{2}-a \right)}{bc}-\frac{\left( \frac{a+b+c}{2}-c \right)\left( \frac{a+b+c}{2}-b \right)}{bc}$
$=\frac{1}{bc}\left( \frac{a+b+c}{2}\left( \frac{a+b+c}{2}-a \right)-\left( \frac{a+b+c}{2}-c \right)\left( \frac{a+b+c}{2}-b \right) \right)$
$ =\frac{1}{bc}\left( \frac{a+b+c}{2}\left( \frac{b+c-a}{2} \right)-\left( \frac{a+b-c}{2} \right)\left( \frac{a+c-b}{2} \right) \right)$
$=\frac{1}{4bc}\left[ \left( (b+c)+a \right)((b+c)-a)-(a+(b-c)(a-(b-c) \right]$
Now let us assume $b+c=x$ and $b-c=y$, so}
\[\begin{align}
  & =\frac{1}{4bc}\left[ \left( x+a \right)(x-a)-(a+y)(a-y) \right] \\
 & =\frac{1}{4bc}\left[ ({{x}^{2}}-{{a}^{2}})-({{a}^{2}}-{{y}^{2}}) \right] \\
 & =\frac{1}{4bc}\left( {{x}^{2}}-{{a}^{2}}-{{a}^{2}}+{{y}^{2}} \right) \\
 & =\frac{1}{4bc}\left( {{x}^{2}}-2{{a}^{2}}+{{y}^{2}} \right) \\
\end{align}\]
We will now substitute back the value of $x$ and $y$.
\[\begin{align}
  & =\frac{1}{4bc}\left( {{(b+c)}^{2}}-2{{a}^{2}}+{{(b-c)}^{2}} \right) \\
 & =\frac{1}{4bc}\left( {{b}^{2}}+{{c}^{2}}+2bc-2{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-2bc \right) \\
 & =\frac{2{{b}^{2}}+2{{c}^{2}}-2{{a}^{2}}}{4bc} \\
 & =\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc} \\
\end{align}\]
Now we know that this is the cosine rule for the angle $A$, so the equation will be reduced to,
$=\cos A$
The value of the equation $\frac{s(s-a)}{bc}-\frac{(s-c)(s-a)}{bc}=$for the triangle $\vartriangle ABC$_{when $2s=a+b+c$, is $\cos A$}. Hence the correct option is (B).
Note:

We could have solved the equation without assuming the values $b+c=x$ _{and $b-c=y$. But to simplify the equation} we opted this method. We must form the equation in a way so that we could simplify it using the expansion formulas easily. We must also have the knowledge cosine rule so that we could comprehend that the derived equation at the end is a formula of $\cos A$.
The formula $2s=a+b+c$ _which can be written as $s=\frac{a+b+c}{2}$ is the formula of the semi perimeter of the triangle.