sin≤ft(2tan^-15/12) is equal to

$\sin\left(2\tan^{-1}\frac{5}{12}\right)$

$=\sin\left(\tan^{-1}\left(\frac{2×\frac{5}{12}}{1-\frac{5^2}{12^2}}\right)\right)$  $[∵2\tan^{-1}x=\tan^{-1}\frac{2x}{1-x^2}]$

$=\sin\left[\tan^{-1}\left[\frac{10/12}{(144-25)/144}\right]\right]$

$=\sin\left(\tan^{-1}[\frac{120}{119}]\right)$

let $\tan^{-1}(\frac{120}{119})=x$

Using pythagoras theorem

$=\sqrt{119^2+120^2}=169$

$⇒x=\sin^{-1}(\frac{120}{169})$

$\sin(\sin^{-1}(\frac{120}{169}))$

$=\frac{120}{169}$