Question

In a triangle ABC, ∠C = 90°, then `(a^2 - b^2)/(a^2 + b^2)` is ______.

विकल्प

• sin (A + B)

• sin (A – B)

• cos (A + B)

• `sin((A - B)/2)`

Answer 1

In a triangle ABC, ∠C = 90°, then `(a^2 - b^2)/(a^2 + b^2)` is sin (A – B).

Explanation:

A + B = 180° – C = 90°

a = 2R sin A, b = 2R sin B, c = 2R sin C

∴ `(a^2 - b^2)/(a^2 + b^2) = (sin^2A - sin^2B)/(sin^2A + sin^2B)`

= `(sin(A + B)sin(A - B))/(sin^2A + sin^2(90^circ - A))`  ...[∵ A + B = 90°]

= `(sin 90^circ sin(A - B))/(sin^2A + cos^2A)`

= sin (A – B)