Evaluate

1. sin(tan inverse sqrt(x^2-2x))

2. tan (sec inverse 3y)

To evaluate these expressions, we need to use the inverse trigonometric functions and fundamental trigonometric identities. Here's how you can solve each expression step-by-step:

1. sin(tan^(-1)(√(x^2-2x)))

First, let's simplify the inner expression, tan^(-1)(√(x^2-2x)).
- We can start by finding the value of (√(x^2-2x)).

To solve |x^2-2x|:
- We notice that the expression inside the square root, x^2-2x, is equal to x(x-2).
- Since we have a square root, we are looking for the positive value of x(x-2).
- So, we set x(x-2) = 0 and solve for x: x = 0 or x = 2.

Now, let's find the value of tan^(-1)(√(x^2-2x)):
- When tan^(-1) is applied to a value, it gives the angle whose tangent is that value.
- Since we are dealing with a positive square root, we look for the positive angle whose tangent equals that value.
- So, we take the angle in the first quadrant (0° < θ < 90°), where the tangent is positive.
- Therefore, tan^(-1)(√(x^2-2x)) = tan^(-1)(√(0^2-2*0)) = tan^(-1)(√0) = tan^(-1)(0) = 0°.

Now, we have sin(0°):
- Remember that the sine function returns the y-coordinate of a point on the unit circle.
- At 0°, the y-coordinate is 0.
- So, sin(0°) = 0.

Therefore, the value of sin(tan^(-1)(√(x^2-2x))) is 0.

2. tan(sec^(-1)(3y))

First, let's simplify the inner expression, sec^(-1)(3y):
- The sec^(-1) function gives the angle whose secant is the given value.
- For sec^(-1)(3y):
- We know that sec(θ) = 1/cos(θ), and sec^(-1)(x) is the inverse of the sec function.
- So, sec(θ) = x is equivalent to cos(θ) = 1/x.
- In our case, cos(θ) = 1/(3y).

Now, let's find the value of cos(θ):
- To solve for θ in the equation cos(θ) = 1/(3y):
- We apply the cosine inverse (cos^(-1)) to both sides of the equation.
- Therefore, θ = cos^(-1)(1/(3y)).

Now, we want to find tan(sec^(-1)(3y)):
- Tan(θ) is equal to the ratio of sin(θ) to cos(θ) on the unit circle.
- So, we need to find sin(θ) and cos(θ).

In this case, we have θ = cos^(-1)(1/(3y)). So, we need to find sin(cos^(-1)(1/(3y))):
- To solve this, we use the Pythagorean identity: sin^2(θ) + cos^2(θ) = 1.
- Since we know cos(θ) = 1/(3y), we can rearrange the equation to solve for sin(θ).
- sin(θ) = √(1 - cos^2(θ)) = √(1 - (1/(3y))^2).

Now, we have sin(cos^(-1)(1/(3y))) = √(1 - (1/(3y))^2).

Finally, we can calculate tan(sec^(-1)(3y)):
- tan(sec^(-1)(3y)) = tan(θ) = sin(θ)/cos(θ), where θ = cos^(-1)(1/(3y)).
- So, tan(sec^(-1)(3y)) = (√(1 - (1/(3y))^2))/(1/(3y)) = (√(1 - 1/(9y^2)))/(1/(3y)).

Therefore, the value of tan(sec^(-1)(3y)) is (√(1 - 1/(9y^2)))/(1/(3y)).