tan^-1(x/√a^2-x^2)=sin^-1(x/a)
To prove that the equation tan^(-1)(x/√(a^2-x^2)) = sin^(-1)(x/a), we need to show that the two sides of the equation are equal.
First, let's start with the right-hand side of the equation: sin^(-1)(x/a).
We know that the sine function is defined as opposite over hypotenuse in a right triangle. So, we can start by drawing a right triangle.
Let's label the angle opposite x as θ. Then, the opposite side is x, and the hypotenuse is a.
Now, using the definition of sine, sin(θ) = opposite/hypotenuse, we have sin(θ) = x/a.
Taking the inverse sine (or arcsin) of both sides to isolate θ, we get θ = sin^(-1)(x/a).
Now, let's move to the left-hand side of the equation: tan^(-1)(x/√(a^2-x^2)).
The tangent function is defined as opposite over adjacent in a right triangle. We can use the same triangle we drew before.
Let's label the adjacent side as b. Then, the opposite side is x, and the hypotenuse is a.
Using the Pythagorean theorem, we have a^2 = b^2 + x^2. Solving for b, we get b = √(a^2 - x^2).
Now, applying the definition of tangent, tan(θ) = opposite/adjacent, we have tan(θ) = x/√(a^2 - x^2).
Taking the inverse tangent (or arctan) of both sides to isolate θ, we get θ = tan^(-1)(x/√(a^2 - x^2)).
Therefore, we have shown that sin^(-1)(x/a) = tan^(-1)(x/√(a^2 - x^2)).
If you plug in any value of x and a, you will find that the equation holds true.
Hence, the equation tan^(-1)(x/√(a^2-x^2)) = sin^(-1)(x/a) is proven.