The question should be to prove tan(sin-1(1/√17)+cos-1(9/√85))=½ (not tan-1).
Let sin(x)=1/√17, cos(y)=9/√85, then we need to prove tan(x+y)=½.
In a right triangle, if the hypotenuse has length √17 and one leg length=1, the other leg has length (√(17-1)=4. So tan(x)=¼.
Similarly, tan(y)=2/9.
tan(x+y)=(tan(x)+tan(y))/(1-tan(x)tan(y))=(1/4+2/9)/(1-(1/4)(2/9))=(17/36)/(34/36)=17/34=½ QED