Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

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

• A. $\sin (A+B)$
• B. $\cos (A+B)$
• C. $\cos (A-B)$
• D. $\sin A+\sin B$

Answer 1

Answer: (A)

$\sin (A+B)$

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin 90°}{c}=K$ [Sine rule]

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{1}{c}=K$

$\left(\frac{a^2+b^2}{a^2-b^2}\right)\left(\sin A\cos B-\cos A\sin B\right)=\frac{a^2+b^2}{a^2-b^2}\left[\frac{a}{c}\cos B-\frac{b}{c}\cos A\right]$

$=\frac{a^2+b^2}{a^2-b^2}\times \left[\frac{2(a^2-b^2)}{2c^2}\right]=\left(\frac{a^2+b^2}{c^2}\right)=\frac{c^2}{c^2}=1$

$\sin (A+B)=\sin A\cos B+\cos B\sin A=\frac{a}{c}\left(\frac{a^2+c^2-b^2}{2ac}\right)+\frac{b}{c}\left(\frac{b^2+c^2-a^2}{2bc}\right)$

$=\frac{a^2+c^2-b^2+b^2+c^2-a^2}{2c^2}$

$=1$

Hence, $\sin (A+B)$ is the correct answer.