In a $\triangle A B C, C=90^{\circ}$. Then, $\frac{a^2-b^2}{a^2+b^2}$ is equal to

In a $\triangle A B C, C=90^{\circ}$. Then, $\frac{a^2-b^2}{a^2+b^2}$ is equal to
  1. $\sin (A+B)$
  2. $\sin (A-B)$
  3. $\cos (A+B)$
  4. $\cos (A-B)$

Solution

$\angle C=90^{\circ}, \frac{a^2-b^2}{a^2+b^2}$ ...(i) $\because \quad \cos c=\frac{a^2+b^2-c^2}{2 a b}=\cos 90^{\circ}$ $\Rightarrow \quad \frac{a^2+b^2-c^2}{2 a b}=0$ $a^2+b^2-c^2=0$ $\Rightarrow \quad c^2=a^2+b^2$ From Eq. (i), we get $\frac{a^2-b^2}{a^2+b^2}=\frac{a^2-b^2}{c^2}$ $\because \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=R$ (constant) $=\frac{(R \sin A)^2-(R \sin B)^2}{(R \sin C)^2}=\frac{\sin ^2 A-\sin ^2 B}{\sin ^2 90^{\circ}}$ $\left(\because \sin 90^{\circ}=1\right)$ $=(\sin A+\sin B)(\sin A-\sin B)$ $=2 \cdot \sin \frac{A+B}{2} \cdot \cos \frac{A-B}{2} \cdot 2 \cos \frac{A+B}{2}$ $\cdot \sin \frac{A-B}{2}$ $=2 \sin \frac{\pi}{4} \cdot \cos \frac{A-B}{2} \cdot 2 \cos \frac{\pi}{4} \cdot \sin \frac{A-B}{2}$ $\because A+B=\pi-C$ $=\pi-\frac{\pi}{2}=\frac{\pi}{2}$ $=2 \cdot 2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \cdot \sin \frac{A-B}{2} \cdot \cos \frac{A-B}{2}$ $=2 \cdot \sin \left(\frac{A-B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right)$ $\because \sin 2 A=2 \sin A \cdot \cos A$ $=\sin \left[2 \times \frac{(A-B)}{2}\right]$ $=\sin (A-B)$ Hence, $\quad \frac{a^2-b^2}{a^2+b^2}=\sin (A-B)$

Asked in: AP EAMCET 2010

Practice more Properties of Triangles questions on Aicharya