Write the expression as an algebraic expression in x for $x > 0$ .
$\sin (\tan^{- 1} x)$

Question

Write the expression as an algebraic expression in x for x&amp;gt;0.  sin⁡(tan−1x)

hysgubwyri3 · Accepted Answer

sin⁡(tan−1x) Now let t=tan−1x ⇒tan⁡(t)=x ⇒sin⁡(t)=x1+x2 (because tan⁡t=perpendicularbase⇒perpendicular=x and base=1then hypotenuse=1+x2 also sin⁡t=perpendicularhypotenuse) ⇒t=sin−1x1+x2 So, tan−1x=sin−1x1+x2 ⇒sin⁡(tan−1x)=sin⁡(sin−1x1+x2) =x1+x2

Ethan Sanders · Accepted Answer

Explanation:  sin2(θ)+cos2(θ)=1  We divide both sides by sin2(θ) so we have  1+cot2(θ)=csc2(θ)  Or, 1+1tan2(θ)=1sin2(θ)  Taking the least common multiple we have  tan2(θ)+1tan2(θ)=1sin2(θ)  Inverting both sides we have  sin2(θ)=tan2(θ)tan2(θ)+1  So we say that θ=arctan⁡(x)  sin2(arctan⁡(x))=tan2(arctan⁡(x))tan2(arctan⁡(x))+1  Knowing that tan⁡(arctan⁡(x))=x  sin2(arctan⁡(x))=x2x2+1  So we take the square root of both sides  sin⁡(arctan⁡(x))=±x2x2+1=±|x|x2+1  Checking the range of the arctangent, we see that during it the sine is always positive so we have  sin⁡(arctan⁡(x))=|x|x2+1

RizerMix · Accepted Answer

Answer: if you were to write the expression shown above as an algebraic equation in x without using trig or inverse trig functions then the correct answer choice would be letter D)  D. x1+x2

Write the expression as an algebraic expression in x for

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