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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))
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make a triangle... the ans will be obvious. it is opp/hyp so it is x/(x^2+1)
Ah a triangle! Thank you!
yes a triangle. label "opposite" side x and adjacent side 1 since the tangent is x = x/1 use pythatoras to find the hypotenuse which will be \[\sqrt{1+x^2}\]
then \[sin(tan^{-1}(x))=\frac{x}{\sqrt{1+x^2}}\]
that being opposite over hypotenuse
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and by the same triangle \[cos(tan^{-1}(x))=\frac{1}{\sqrt{1+x^2}}\]
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