Question

In \(△ABC\), with the usual notations, \(\angle C = 90^{\circ}\), then \(sin(A-B)\) is equal to....

• A. \(\frac{a^2+b^2}{a^2-b^2}\)
• B. \(\frac{a^2+c^2}{a^2-c^2}\)
• C. \(\frac{b^2+c^2}{b^2-c^2}\)
• D. \(\frac{a^2-b^2}{a^2+b^2}\)

Hint:

Use sinA = a/c, cosA = b/c and c^2 = a^2 + b^2.

Answer 1

The Correct Option is D

Step 1: Use the angle relation:
$A+B=90^\circ$, so $B=90^\circ-A$. Then $A-B = 2A-90^\circ$.

Step 2: Reduce:
$\sin(2A-90^\circ)=-\cos 2A = \sin^2A-\cos^2A$.

Step 3: Put the sides:
$\sin^2A = a^2/c^2$ and $\cos^2A = b^2/c^2$. So result $= (a^2-b^2)/c^2 = (a^2-b^2)/(a^2+b^2)$.

Final Answer:
Both methods give (a^2 - b^2)/(a^2 + b^2). \[ \boxed{D} \]