Question

In $ABC$, if $C={90}^o$ then $\frac{a^2+b^2}{a^2-b^2}\sin$ ($A-B$) $=$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

Consider, $\frac{a^2+b^2}{a^2-b^2}\sin (A-B)$
$=\frac{k^2({\sin }^{2}A+{\sin }^{2}B)}{k^2({\sin }^{2}A-{\sin }^{2}B)}\sin (A-B)$
$=\frac{({\sin }^{2}A+{\sin }^{2}B)}{\sin (A+B)\sin (A-B)}\sin (A-B)$ ....... $[\because \sin (A+B\left)\sin \right(A-B)={\sin }^{2}A-{\sin }^{2}B]$
$=\frac{({\sin }^{2}A+{\sin }^{2}B)}{\sin (A+B)}$
$={\sin }^{2}B+1-{\cos }^{2}A$
$=1-({\cos }^{2}A-{\sin }^{2}B)$ ....... $[\because \angle C={90}^o\Rightarrow A+B={90}^o]$
$=1-\cos (A+B)\cos (A-B)$ ........ $[{\cos }^{2}A-{\sin }^{2}B=\cos (A+B)\cos (A-B)]$
$=1-0=1$ ....... $[\because \angle C={90}^0\Rightarrow A+B={90}^o]$