Write the trigonometric expression sin(sin−1u−tan−1v) as an algebraic expression in u and v. Assume that the variables u and v represent positive real numbers.

[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).

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step-by-step explanation:

i think the answer is true

step-by-step explanation: