posted by David M. .
Usa a right triangle to write each expression as an algebraic expression. Assume the referenced angle is theta, and x is positive and in the domain of the given inverse trig. function. Label the triangle.
2. sin(arccot x/(sqrt)1-x^2)
1. If you mean sec^2[tan^-1x], use the fact that
sec (tan^-1x)= sqrt(x^2+1)/sqrt(1).
Imagine a right triangle with sides x, 1 and sqrt(1+x^2). Side x is opposite the angle with tangent equal to x.
Square the secant and you get x^2 + 1
2. For the angle in question, as part of a right triangle:
adjacent side (not hypotentuse) = x
opposite side = sqrt(1-x^2)
hypotenuse = 1
sin = sqrt(1-x^2)
The last line is your answer.