Answer:
[u – v√(1 – u²)]/√(1 + v²)
Step-by-step explanation:
Let sin^-1(u) = A, therefore sinA = u.
We know that sin(theta) = opposite/hypothenuse
Therefore, sinA = u/1 and u is the opposite side to angle A while 1 is the hypotenuse. Draw an acute triangle placing u opposite to angle A and 1 as the hypotenuse. By Pythagoras theorem the adjacent would be √(1 – u²).
By doing this, it means cosA = adjacent/hypotenuse = √(1 – u²)/1 = √(1 – u²)
Also, let tan^-1(v) = B, therefore tanB = v.
We know that tan(theta) = opposite/adjacent
Therefore, tanB = v/1 and v is the opposite side to angle B while 1 is the adjacent. Draw an acute triangle placing v opposite to angle B and 1 as the adjacent. By Pythagoras theorem the hypothenuse would be √(1 + v²).
Therefore, sinB = opposite/hypotenuse = v/√(1 + v²) and cosB = adjacent/hypotenuse = 1/√(1 + v²)
Now,
sin[sin^–1(u) – tan^–1(v)] =
sin(A – B) =
sinAcosB – sinBcosA =
u[1/√(1 + v²)] – [v/√(1 + v²)][√(1 – u²)] =
[u/√(1 + v²)] – [v√(1 – u²)/√1 + v²)] =
[u – v√(1 – u²)]/√(1 + v²).
