# How do you evaluate sin(tan^-1(12/5))?

$\sin \left({\tan}^{-} 1 \left(\frac{12}{5}\right)\right) = \frac{12}{13}$
$\sin \left({\tan}^{-} 1 \left(\frac{12}{5}\right)\right)$. Let ${\tan}^{-} 1 \left(\frac{12}{5}\right) = \theta \therefore \tan \theta = \frac{12}{5}$
Since ${\tan}^{-} 1$ exists in 1st quadrant & 4th quadrant and positive in 1st quadrant, $\theta$ is in 1st quadrant . we know tan theta =p/b= 12/5 :. p=12 ; b=5 ; h= sqrt(p^2+b^2)=sqrt(12^2+5^2)=13 ; sin theta = p/h=12/13. [p=perpendicular ; b = base; h= hypotenuse]
$\therefore \sin \left({\tan}^{-} 1 \left(\frac{12}{5}\right)\right) = \sin \theta = \frac{12}{13}$ [Ans]