let assume
[tex]sin( {tan}^{ - 1} 5) = sin \: y \\ so \: y \: = {tan}^{ - 1} 5 \: \\ tan \: y \: = 5 \\ [/tex]
we know from defination of inverse function
tan y = opposite / adjacent
here opposite is 5 and adjacent is 1
tan y = 5/1
lets calculate hypotenus using pythagoras theorem
hypotenuse = root ( 5^2 + 1^2)= root (25+1) = root (26)
now we know
sin y = opposite / hypotenus
so
[tex]sin \: y \: = \frac{5}{ \sqrt{26} } \\ sin \: ( {tan}^{ - 1} x)\: = \frac{5}{ \sqrt{26} } [/tex]
hope it helps you
