Question

In \(\Delta ABC\), if \(\angle A = 90^{\circ}\), then \(sin(B-C) =\)

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

Hint:

Use sinB = b/a, cosB = c/a, sinC = c/a, cosC = b/a and a^2 = b^2 + c^2.

Answer 1

The Correct Option is C

Step 1: Use B + C = 90 degrees:
Then $C=90^{\circ}-B$, so $B-C=2B-90^{\circ}$.

Step 2: Simplify:
$\sin(2B-90^{\circ})=-\cos 2B = \sin^2B-\cos^2B$.

Step 3: Substitute:
$\sin B=b/a$, $\cos B=c/a$: $\dfrac{b^2-c^2}{a^2}=\dfrac{b^2-c^2}{b^2+c^2}$. Option C.

Final Answer:
Using B + C = 90 degrees gives (b^2 - c^2)/(b^2 + c^2). \[ \boxed{\text{(C) }\dfrac{b^2-c^2}{b^2+c^2}} \]