Help Please!

Simplify $\sin \left( \sin^{-1} \frac{3}{5} + \tan^{-1} 2 \right).$

Hello Guest!

$sin(\frac{1}{sin(\frac{3}{5})}+\frac{1}{tan(2)})\\ =sin(csc(\frac{3}{5})+\frac{1}{tan(2)})\\ =sin(\pm \frac{\sqrt{1+tan^2(\frac{3}{5})}}{tan(\frac{3}{5})}+\frac{1}{tan(2)})$

I dont get any further.

  !

Assuming $\sin^{-1} x$ and $\tan^{-1} x$ means inverse trigonometric functions,

$\sin\left(\sin^{-1} \dfrac35 + \tan^{-1} 2\right)\\ = \sin\left(\tan^{-1} \dfrac34 + \tan^{-1}2 \right)\\ = \sin \tan^{-1} \dfrac34 \cos \tan^{-1} 2 + \cos \tan^{-1} \dfrac34 \sin \tan^{-1} 2\\ = \dfrac35 \dfrac1{\sqrt5} + \dfrac45\dfrac{2}{\sqrt5}\\ = \dfrac{11}{5\sqrt5}$