If sin-1 x + sin-1 y + sin-1 z = π, then prove that x√1 - x2 + y√1 - y2 + z√1 - z2 = 2xyz
Solution:
Inverse trigonometric functions are the inverse ratio of the basic trigonometric ratios.
Given: sin-1 x + sin-1 y + sin-1 z = π ---------------- (1)
To prove: x√1 - x2 + y√1 - y2 + z√1 - z2 = 2xyz
Let, sin-1 x = A, sin-1 y = B, sin-1 z = C
Thus,
x = sin A --------------- (2)
y = sin B --------------- (3)
z = sin C --------------- (4)
Substituting the values of sin-1 x, sin-1 y, and sin-1 z in (1)
⇒ A + B + C = π
Now, let's take the LHS x√1 - x2 + y√1 - y2 + z√1 - z2
Substituting the values of x, y and z from (2), (3) and (4) we get,
⇒ sin A√1 - sin2 A + sin B√1 - sin2 B + sin C√1 - sin2 C
⇒ sinA.cos A + sin B.cos B + sin C.cos C
Multiplying and dividing by 2 we get,
⇒ 2 / 2 [sin A.cos A + sin B.cos B] + sin C.cos C
⇒ (2sin A.cos A + 2sin B.cos B) / 2 + sin C.cos C
⇒ (sin 2A + sin 2B) / 2 + sin C.cos C
Since, 2sin θ.cos θ = sin 2θ
⇒ [2sin[(2A + 2B) / 2].cos[(2A − 2B) / 2]] / 2 + sin C.cos C
⇒ sin(A + B).cos(A − B) + sin C.cos C
⇒ sin(π − C).cos(A − B) + sinC.cos(π − (A + B))
⇒ sin C [cos(A − B) − cos(A + B)]
Since, sin(π − x) = sin x, cos(π − x) = - cos x
⇒ sinC [−2sin (-2B/2) sin(2A/2)]
⇒ 2sin A.sin B.sin C
⇒ 2xyz [From (2), (3) and (4)] = RHS
Hence, proved x√1 - x2 + y√1 - y2 + z√1 - z2 = 2xyz
If sin-1 x + sin-1 y + sin-1 z = π, then prove that x√1 - x2 + y√1 - y2 + z√1 - z2 = 2xyz
Summary:
If sin-1 x + sin-1 y + sin-1 z = π, its proved that x√1 - x2 + y√1 - y2 + z√1 - z2 = 2xyz
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