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: Understanding the Concept:
In a right triangle with \(\angle C=90^\circ\), the hypotenuse is \(c\), so \(a^2+b^2=c^2\), \(\sin A = a/c\), \(\cos A = b/c\), \(\sin B = b/c\) and \(\cos B = a/c\).

Step 2: Expand:
\[ \sin(A-B)=\sin A\cos B-\cos A\sin B \]

Step 3: Substitute:
\[ \sin(A-B)=\frac{a}{c}\cdot\frac{a}{c}-\frac{b}{c}\cdot\frac{b}{c}=\frac{a^2-b^2}{c^2} \]

Step 4: Replace c squared:
Since \(c^2=a^2+b^2\), we get \(\sin(A-B)=\dfrac{a^2-b^2}{a^2+b^2}\), option (D).

Final Answer:
sin(A-B) = (a^2 - b^2)/(a^2 + b^2). \[ \boxed{\frac{a^2-b^2}{a^2+b^2}} \]