Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (A)

$1$

Given that, In $\Delta ABC$,$\angle C={90}^0$

$A+B+C={180}^0$

$A+B={180}^0-{90}^0$

$A+B={90}^0........\left(1\right)$

Then,

$\frac{a^2+b^2}{a^2-b^2}\sin (A-B)$

Using sine rule, and we get,

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

$\frac{{\left(k\sin A\right)}^2+{\left(k\mathrm{sinB}\right)}^2}{{\left(k\sin A\right)}^2-{\left(k\mathrm{sinB}\right)}^2}\sin (A-B)$

$=\frac{{\sin }^{2}A+{\sin }^{2}B}{{\sin }^{2}A-{\sin }^{2}B}\sin (A-B)$

$=\frac{\frac{1-\cos 2A}{2}+\frac{1-\cos 2B}{2}}{\frac{1-\cos 2A}{2}-\frac{1-\cos 2B}{2}}\sin (A-B)\therefore \cos 2\theta =1-2{\sin }^2\theta$

$=\frac{1-\cos 2A+1-\cos 2B}{1-\cos 2A-1+\cos 2B}\sin (A-B)$

$=\frac{2-\cos 2A-\cos 2B}{\cos 2B-\cos 2A}\sin (A-B)$

$=\frac{2-(\cos 2A+\cos 2B)}{\cos 2B-\cos 2A}\sin (A-B)$

$=\frac{2-\left(2\cos \frac{(2A+2B)}{2}\cos \frac{(2A-2B)}{2}\right)}{2sin\frac{(2B+2A)}{2}\sin \frac{(2A-2B)}{2}}\sin (A-B)$

$=\frac{2-\left(2\cos \right(A+B\left)\cos (A-B)\right)}{2\sin (A+B)\sin (A-B)}\sin (A-B)$

$=\frac{2-\left(2\cos \right(A+B\left)\cos (A-B)\right)}{2\sin (A+B)\sin (A-B)}\sin (A-B)$

$=\frac{2-\left(2\cos \right(A+B\left)\cos (A-B)\right)}{2\sin (A+B)}$

Now, by equation (1) and we get,

$=\frac{2-\left(2\cos {90}^0\cos (A-B)\right)}{2\sin {90}^0}$

$=\frac{2-0}{2}$

$=1$

Hence, this is the answer.