Please, rewrite the expression as an algebraic expression in x.
$\sin (\tan^{-} 1 x)$

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Please, rewrite the expression as an algebraic expression in x. sin⁡(tan−⁡1x)

sweererlirumeX · Accepted Answer

Let θ=tan−1(x), then tan⁡θ=x Draw a right triangle with θ as one of it’s acute angles: 1) AC=AB2+DC2 AC=x2+12=1+x2 2) sin⁡(tan−1(x))=x1+x2

karton · Accepted Answer

Step 1: Recall the identity tan−1x=arctan(x).Step 2: Use the definition of the tangent function: tan(θ)=sin(θ)cos(θ). In this case, θ=tan−1x. So we can rewrite tan−1x as arctan(x)=sin(arctan(x))cos(arctan(x)).Step 3: Apply the identity sin(arctan(x))=x1+x2. This can be derived using the right triangle definition of the trigonometric functions.Step 4: Apply the identity cos(arctan(x))=11+x2. Again, this can be derived using the right triangle definition of the trigonometric functions.Step 5: Substitute the values of sin(arctan(x)) and cos(arctan(x)) back into the expression sin(arctan(x))cos(arctan(x)).Combining all the steps, we get:sin(tan−1x)=sin(sin(arctan(x))cos(arctan(x)))&amp;nbsp;=sin(arctan(x))cos(arctan(x))&amp;nbsp;=x1+x211+x2&amp;nbsp;=x1+x2·1+x2&amp;nbsp;=x·1+x21+x2&amp;nbsp;=xTherefore, the expression sin(tan−1x) can be rewritten as the algebraic expression x.

star233 · Accepted Answer

To rewrite the expression sin(tan−1x) as an algebraic expression in x, we can use trigonometric identities. First, we&#039;ll start by using the identity tan−1x=arctan(x):sin(tan−1x)=sin(arctan(x))Next, we can use the identity sin(arctan(x))=x1+x2:sin(tan−1x)=sin(arctan(x))=x1+x2Therefore, the algebraic expression in x equivalent to sin(tan−1x) is x1+x2.

user_27qwe · Accepted Answer

Answer:x1+x2Explanation:tan−1x=arctanxUsing this identity, we have:sin(tan−1x)=sin(arctanx)Now, let&#039;s use another trigonometric identity to express sin(arctanx) in terms of x. The identity is:sin(arctanx)=x1+x2Therefore, the algebraic expression in x equivalent to sin(tan−1x) is:x1+x2

Please, rewrite the expression as an algebraic expression in x. sin(tan^

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