Write the expression sin(tan^-1)x) as an algebraic expression in x (without trig or inverse trig functions).

a. (sqrt (1+x^2))/ (x)
b. (1) / (sqrt (1+x^2))
c. (x) / (sqrt (1-x^2))
d. (x) / (sqrt (1+x^2))
e. (1) / (sqrt (1-x^2))

Option “D” (x) / (sqrt (1+x^2)) is correct.

As sin^2(α) + cos^2(α) = 1

By division of sin^2 (α) on both sides of equation, we get

1 + 1/tan^2(α) = 1/sin^2(α)

By taking L.C.M and inverting the whole equation;

Sin^2(α) = tan^2(α) / (tan^2(α) + 1)

Let α = arctan(x);

Sin^2 (arctan(x)) = tan^2 (arctan(x)) / (tan^2(arctan(x)) +1)

As tan(arctan(x)) = x

Hence

Sin^2 (arctan(x)) = x^2 / (x^2 + 1)

Now taking square root;

Sin (arctan(x)) = (x) / (sqrt (1+x^2))

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step-by-step explanation: