In A B C, if A=90^{}, then (B-C)=

Correct Answer is : b2−c2b2+c2{b²-c²}{b²+c²}b2+c2b2−c2

Correct Answer is : b2−c2b2+c2{b²-c²}{b²+c²}b2+c2b2−c2

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MHTCET 11 Apr 2026 Shift 2 Paper

Question : 103 of 150

Marks: +1, -0

\triangle A B C

\angle A=90^{\circ}

\sin (B-C)=

Solution:

Concept:

In a right triangle, express trigonometric ratios using sides, then simplify

\sin(B-C)

using the identity.

Explanation:

\triangle ABC

, side

a

is opposite

\angle A = 90^\circ

a

is the hypotenuse, and

b

c

are the other two sides.

Since

B + C = 90^\circ

, we can directly use:

\sin(B-C) = \sin B \cos C - \cos B \sin C

For the right triangle:

\sin B = \frac{b}{a}, \quad \cos C = \frac{b}{a}

\cos B = \frac{c}{a}, \quad \sin C = \frac{c}{a}

Substitute these values:

\sin(B-C) = \frac{b}{a}\cdot\frac{b}{a} - \frac{c}{a}\cdot\frac{c}{a}

\sin(B-C) = \frac{b^2 - c^2}{a^2}

By Pythagoras theorem,

a^2 = b^2 + c^2

Hence:

\sin(B-C) = \frac{b^2 - c^2}{b^2 + c^2}

Answer:

\frac{b^{2}-c^{2}}{b^{2}+c^{2}}

Therefore, the correct option is

\text{Option C}

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