Question

In a ∆ABC, if $\frac{\sin \left(A–B\right)}{\sin \left(A+B\right)}=\frac{a^2-b^2}{k},$ then k = ___________.

Answer 1

In ∆ABC,

Given $\frac{\sin \left(A-B\right)}{\sin \left(A+B\right)}=\frac{\sin A\cos B-\cos A\sin B}{\sin A\cos B+\cos A\sin B}$

$$\begin{gathered} \mathrm{Since}\cos A\mathit{}\mathit{=}\mathit{}\frac{b^{\mathit{2}}+\mathit{}{c}^{\mathit{2}}\mathit{-}\mathit{}{a}^{\mathit{2}}}{\mathit{2}bc} \\ \cos B=\mathit{}\frac{a^{\mathit{2}}+\mathit{}{c}^{\mathit{2}}-\mathit{}{b}^{\mathit{2}}}{\mathit{2}ac} \\ \mathrm{and}\mathit{}\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=\mathit{}x\mathit{}\mathrm{say}\mathit{} \end{gathered}$$
$\begin{array}{rcl}\mathrm{i}.\mathrm{e}\mathit{}\frac{\sin \mathit{}(A-B)}{\sin (A+B)}& =& \frac{\left({\displaystyle \frac{\mathrm{a}}{\mathrm{x}}}\right)\left(\frac{a^{\mathit{2}}+c^{\mathit{2}}-b^{\mathit{2}}}{\mathit{2}ac}\right)\mathit{-}\left(\frac{b^{\mathit{2}}+c^{\mathit{2}}-a^{\mathit{2}}}{\mathit{2}bc}\right)\left(\frac{b}{x}\right)}{{\displaystyle \frac{a}{x}\left(\frac{a^{\mathit{2}}+c^{\mathit{2}}-b^{\mathit{2}}}{\mathit{2}ac}\right)+\left(\frac{b^{\mathit{2}}+c^{\mathit{2}}-a^{\mathit{2}}}{\mathit{2}bc}\right)\left(\frac{b}{x}\right)}}\\ & =& \frac{{\displaystyle \frac{a^{\mathit{2}}+c^{\mathit{2}}-b^{\mathit{2}}}{\mathit{2}cx}\mathit{}\mathit{-}\mathit{}\mathit{}\frac{b^{\mathit{2}}+c^{\mathit{2}}-a^{\mathit{2}}}{\mathit{2}cx}}}{{\displaystyle \frac{a^{\mathit{2}}+c^{\mathit{2}}-b^{\mathit{2}}}{\mathit{2}cx}+\mathit{}\frac{b^{\mathit{2}}+c^{\mathit{2}}-a^{\mathit{2}}}{\mathit{2}cx}}}\\ & =& \frac{a^{\mathit{2}}+c^{\mathit{2}}-b^{\mathit{2}}\mathit{-}{b}^{\mathit{2}}-c^{\mathit{2}}+a^{\mathit{2}}}{a^{\mathit{2}}+c^{\mathit{2}}-b^{\mathit{2}}+b^{\mathit{2}}+c^{\mathit{2}}-a^{\mathit{2}}}\\ & =& \frac{\mathit{2}\mathit{}{a}^{\mathit{2}}-\mathit{2}{b}^{\mathit{2}}}{\mathit{2}{c}^{\mathit{2}}}=\frac{a^{\mathit{2}}\mathit{-}{b}^{\mathit{2}}}{c^{\mathit{2}}}=\frac{a^{\mathit{2}}-b^{\mathit{2}}}{k}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\mathrm{given}\right)\\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathrm{i}.\mathrm{e}.k\mathit{}& \mathit{=}& \mathit{}{c}^{\mathit{2}}\end{array}$