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: Understanding the Concept:
When \(\angle A=90^{\circ}\), side \(a\) is the hypotenuse and \(a^2=b^2+c^2\).

Step 2: Write the ratios:
\(\sin B=\dfrac{b}{a},\ \cos B=\dfrac{c}{a},\ \sin C=\dfrac{c}{a},\ \cos C=\dfrac{b}{a}\).

Step 3: Expand sin(B-C):
\[ \sin(B-C)=\sin B\cos C-\cos B\sin C = \dfrac{b}{a}\cdot\dfrac{b}{a}-\dfrac{c}{a}\cdot\dfrac{c}{a}=\dfrac{b^2-c^2}{a^2} \]

Step 4: Replace a squared:
Since \(a^2=b^2+c^2\), \(\sin(B-C)=\dfrac{b^2-c^2}{b^2+c^2}\). This is option C.

Step 5: Why the other options are wrong.
Options A and B have \(a^2\) and \(c^2\) in the numerator, and D has the wrong sign (it equals \(\sin(C-B)\)).

Final Answer:
sin(B-C) equals (b^2 - c^2)/(b^2 + c^2). \[ \boxed{\text{(C) }\dfrac{b^2-c^2}{b^2+c^2}} \]