Question

In a triangle $ABC,\angle C=90^{\circ }$, then $\frac{a^2-b^2}{a^2+b^2}$ is equal to :

• A. $\sin (A+B)$
• B. $\sin (A-B)$
• C. $\cos (A+B)$
• D. $\sin \left(\frac{A-B}{2}\right)$

Answer 1

Answer: (B)

$A+B=180^{\circ }-C=90^{\circ }$
$a=2R\sin A,$
$b=2R\sin B,$
$c=2R\sin C$
$\therefore \frac{a^2-b^2}{a^2+b^2}=\frac{{\sin }^{2}A-{\sin }^{2}B}{{\sin }^{2}A+{\sin }^{2}B}$
$=\frac{\sin (A+B)\sin (A-B)}{{\sin }^{2}A+{\sin }^2\left(90^{\circ }-A\right)}$
$\left[\because A+B=90^{\circ }\right]$
$=\frac{\sin 90^{\circ }\sin (A-B)}{{\sin }^{2}A+{\cos }^{2}A}$
$=\sin (A-B)$