Q66 / 150 · MHT CET 2026 (11 April Shift 2) · Mathematics
In $\Delta ABC$, if $\angle A = 90^\circ$, then $\sin(B - C) = $
- A$\dfrac{a^2 - c^2}{a^2 + c^2}$
- B$\dfrac{c^2 - a^2}{c^2 + a^2}$
- Correct answerC$\dfrac{b^2 - c^2}{b^2 + c^2}$
- D$\dfrac{c^2 - b^2}{c^2 + b^2}$
Solution
Given $\angle A = 90^\circ$, $\Delta ABC$ is a right-angled triangle with hypotenuse $a$.
From the properties of a right-angled triangle, $\sin B = \dfrac{b}{a}$, $\cos B = \dfrac{c}{a}$, $\sin C = \dfrac{c}{a}$, and $\cos C = \dfrac{b}{a}$.
Using the expansion formula for sine:
$\sin(B - C) = \sin B \cos C - \cos B \sin C$
Substituting the trigonometric ratios:
$\sin(B - C) = \left(\dfrac{b}{a}\right)\left(\dfrac{b}{a}\right) - \left(\dfrac{c}{a}\right)\left(\dfrac{c}{a}\right)$
$\sin(B - C) = \dfrac{b^2 - c^2}{a^2}$
Using Pythagoras theorem, $a^2 = b^2 + c^2$.
Therefore, $\sin(B - C) = \dfrac{b^2 - c^2}{b^2 + c^2}$.
From the properties of a right-angled triangle, $\sin B = \dfrac{b}{a}$, $\cos B = \dfrac{c}{a}$, $\sin C = \dfrac{c}{a}$, and $\cos C = \dfrac{b}{a}$.
Using the expansion formula for sine:
$\sin(B - C) = \sin B \cos C - \cos B \sin C$
Substituting the trigonometric ratios:
$\sin(B - C) = \left(\dfrac{b}{a}\right)\left(\dfrac{b}{a}\right) - \left(\dfrac{c}{a}\right)\left(\dfrac{c}{a}\right)$
$\sin(B - C) = \dfrac{b^2 - c^2}{a^2}$
Using Pythagoras theorem, $a^2 = b^2 + c^2$.
Therefore, $\sin(B - C) = \dfrac{b^2 - c^2}{b^2 + c^2}$.
Hence, the correct answer is option: 'C'.