In triangle $\mathrm{ABC}$, if $\mathrm{A}=\frac{\pi}{3}$ and $\mathrm{B}=\frac{\pi}{4}$ then $\frac{a^2-b^2}{c^2}=$

• Correct answer
  A

  $2-\sqrt{3}$
• B

  $2+\sqrt{3}$
• C

  $\sqrt{2}-1$
• D

  $\sqrt{2}+1$

Solution

Given angle $\mathrm{A}=\frac{\pi}{3}$ and $\mathrm{B}=\frac{\pi}{4}$ In $\triangle \mathrm{ABC}$, $$ \begin{aligned} & \angle \mathrm{C}=\pi-(\mathrm{A}+\mathrm{B}) \\ & =\pi-\left(\frac{\pi}{3}+\frac{\pi}{4}\right) \\ & =\pi-\frac{7 \pi}{12}=\frac{5 \pi}{12} \end{aligned} $$ Apply sine law, $$ \begin{aligned} & \frac{\sin \mathrm{A}}{\mathrm{a}}=\frac{\sin \mathrm{B}}{\mathrm{b}}=\frac{\sin \mathrm{C}}{\mathrm{c}} \\ & \frac{\sin \left(\frac{\pi}{3}\right)}{\mathrm{a}}=\frac{\sin \left(\frac{\pi}{4}\right)}{\mathrm{b}}=\frac{\sin \left(\frac{5 \pi}{12}\right)}{\mathrm{c}} \\ & \frac{\sqrt{3}}{2 \mathrm{a}}=\frac{1}{\sqrt{2} \mathrm{~b}}=\frac{\sqrt{3}+1}{2 \sqrt{2} \mathrm{c}} \\ & \text { Compare } \frac{\sqrt{3}}{2 \mathrm{a}}=\frac{\sqrt{3}+1}{2 \sqrt{2} \mathrm{c}} \end{aligned} $$
Similarly, $\frac{1}{\sqrt{2} \mathrm{~b}}=\frac{\sqrt{3}+1}{2 \sqrt{2} \mathrm{c}}$
$$ \begin{aligned} & \text { Take, } \frac{a^2}{c^2}-\frac{b^2}{c^2}=\frac{6}{(\sqrt{3}+1)^2}-\frac{4}{(\sqrt{3}+1)^2} \\ & \frac{2}{(3+1+2 \sqrt{3})}=\frac{2}{4+2 \sqrt{3}}=\frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}} \\ & \frac{a^2-b^2}{c^2}=2-\sqrt{3} \end{aligned} $$ So, correct option is (a).

Hence, the correct answer is option: 'A'.