Question

if in triangle$\vartriangle ABC$_{,$\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=\frac{\sin (A-B)}{\sin (A+B)}$ then the triangle is}
A. Right angled triangle.
B. Isosceles
C. Right angled or isosceles.
D. Right angled isosceles.

Answer 1

Hint: To solve this question, we will first take the given equation and reciprocate it. After this we will apply componendo and dividendo and simplify it using trigonometric formulas. After this we will use sine Law $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$and cosine rules and deduce an equation which in terms will state the type of the triangle.
Formula used:
The cosine rules are:
\[\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}=\cos A\]
\[\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}=\cos B\]
$\begin{align}
  & \sin (A+B)+\sin (A-B)=2\sin A\cos B \\
 & \sin (A+B)-\sin (A-B)=2\cos A\sin B \\
\end{align}$
Complete step-by-step solution:
We are given a triangle $\vartriangle ABC$ <sub>for which $\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=\frac{\sin (A-B)}{\sin (A+B)}$ and we have to determine the type of the triangle.
We will take the given equation $\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=\frac{\sin (A-B)}{\sin (A+B)}$ and then reciprocate it first.
$\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}-{{b}^{2}}}=\frac{\sin (A+B)}{\sin (A-B)}$
Now we will use componendo and dividendo on both the sides,

$\frac{{{a}^{2}}+{{b}^{2}}+{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}-({{a}^{2}}-{{b}^{2}})}=\frac{\sin (A+B)+\sin (A-B)}{\sin (A+B)-\sin (A-B)}$
$\frac{2{{a}^{2}}}{2{{b}^{2}}}=\frac{\sin (A+B)+\sin (A-B)}{\sin (A+B)-\sin (A-B)}$
$\frac{{{a}^{2}}}{{{b}^{2}}}=\frac{\sin (A+B)+\sin (A-B)}{\sin (A+B)-\sin (A-B)}$
We will now use the formula of
$\sin (A+B)+\sin (A-B)$and $\sin (A+B)-\sin (A-B)$</sub>.
$\begin{align}
  & \frac{{{a}^{2}}}{{{b}^{2}}}=\frac{2\sin A\cos B}{2\cos A\sin B} \\
 & \frac{{{a}^{2}}}{{{b}^{2}}}=\frac{\sin A\cos B}{\cos A\sin B}
\end{align}$
Now we will take sine law$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$ and equate it to some constant $\frac{1}{k}$. So the value of angles $\sin A$and $\sin B$will be,
$\begin{align}
  & \sin A=ak \\
 & \sin B=bk \\
\end{align}$
We will now substitute the value of angles $\sin A$and $\sin B$in $\frac{{{a}^{2}}}{{{b}^{2}}}=\frac{\sin A\cos B}{\cos A\sin B}$.
$\begin{align}
  & \frac{{{a}^{2}}}{{{b}^{2}}}=\frac{ak\cos B}{bk\cos A} \\
 & \frac{a}{b}=\frac{\cos B}{\cos A} \\
\end{align}$
We will now use cosine rule for the angle $\cos A$ and $\cos B$ and substitute it.
\[\frac{a}{b}=\frac{\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}}{\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}}\]
\[\frac{a}{b}=\frac{\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{a}}{\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{b}}\]
We will now simplify the equation,
\[\begin{align}
  & \frac{a}{b}=\frac{\left( {{a}^{2}}+{{c}^{2}}-{{b}^{2}} \right)b}{\left( {{b}^{2}}+{{c}^{2}}-{{a}^{2}} \right)a} \\
 & \left( {{b}^{2}}+{{c}^{2}}-{{a}^{2}} \right){{a}^{2}}=\left( {{a}^{2}}+{{c}^{2}}-{{b}^{2}} \right){{b}^{2}} \\
 & {{a}^{2}}{{b}^{2}}+{{a}^{2}}{{c}^{2}}-{{a}^{4}}={{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}-{{b}^{4}} \\
 & {{a}^{2}}{{c}^{2}}-{{b}^{2}}{{c}^{2}}={{a}^{4}}-{{b}^{4}} \\
 & {{c}^{2}}({{a}^{2}}-{{b}^{2}})=({{a}^{2}}-{{b}^{2}})({{a}^{2}}+{{b}^{2}}) \\
 & {{c}^{2}}({{a}^{2}}-{{b}^{2}})-({{a}^{2}}-{{b}^{2}})({{a}^{2}}+{{b}^{2}})=0 \\
 & ({{a}^{2}}-{{b}^{2}})\left[ {{c}^{2}}-({{a}^{2}}+{{b}^{2}}) \right]=0
\end{align}\]
We will now equate both equation to zero.
\[({{a}^{2}}-{{b}^{2}})=0\] or \[\left[ {{c}^{2}}-({{a}^{2}}+{{b}^{2}}) \right]=0\]
$a=b$ or ${{c}^{2}}={{a}^{2}}+{{b}^{2}}$
We know that $a=b$is the condition of an isosceles triangle and ${{c}^{2}}={{a}^{2}}+{{b}^{2}}$ is the condition for a right angled triangle.
The triangle $\vartriangle ABC$ for which $\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=\frac{\sin (A-B)}{\sin (A+B)}$_{, the triangle is either an isosceles triangle or a} right angled triangle. Hence the correct option is (C).
Note:
The componendo and dividendo rule states that if \[\frac{a}{b}=\frac{c}{d}\] then $\frac{a+b}{a-b}=\frac{c+d}{c-d}$.
In a right angled triangle, the relationship between the three sides $a,b,c$of the triangle is ${{c}^{2}}={{a}^{2}}+{{b}^{2}}$.