Question

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