Show that `sin^(-1)(12)/(13)+cos^(-1)4/5+tan^(-1)(63)/(16)=pi`.

Explanation

Let `sin^(-1)((12)/(13))=x and cos^(-1)((4)/(5))=y`
Then `sinx=(12)/(13),"where" 0 lt x lt (pi)/(2)`
and `cosy=(4)/(5), "where" 0lt y lt (pi)/(2)`
`therefore cos x gt 0 and sin y gt 0`
Now , `cos x=sqrt(1-sin^2x)=sqrt((1-144)/(169))=sqrt((25)/(169))=(5)/(13)`
and `sin y =sqrt(1-cos^2y)=sqrt(1-(16)/(25))=sqrt((9)/(25))=(3)/(5)`
`thereforetanx=(sinx)/(cosx)=((12//13))/((5//13))=(12)/(5)`
and `tany=(siny)/(cosy)=((3//5))/((4//5))=(3)/(4)`
`therefore x=tan^(-1)((12)/(5)) and y=tan^(-1)((3)/(4))`
`therefore sin^(-1)((12)/(13))=tan^(-1)((12)/(5))and cos^(-1)((4)/(5))=tan^(-1)((3)/(4))`
`therefore` LHS `sin^(-1)((12)/(13))+cos^(-1)((4)/(5))+tan^(-1)((63)/(16))`
`=tan^(-1)((12)/(5))+tan^(-1)((3)/(4))+tan^(-1)((63)/(16))`
`=pi +tan^(-1)[((12)/(5)+(3)/(4))/(1-(12)/(5)xx(3)/(4))]+tan^(-1)((63)/(16))....[because tan^(-1)x+tan^(-1)y=pi+tan^(-1)((x+y)/(1-xy)),if x gt 0,ygt0,xygt1]`
`=pi+tan^(-1)((48+15)/(20-36))+tan^(-1)((63)/(16))`
`=pi +tan^(-1)(-(63)/(16))+tan^(-1)((63)/(16))`
`=pi-tan^(-1)((63)/(16))+tan^(-1)((63)/(16))...[because tan^(-1)(-theta)=-tan^(-1)theta]`
`= pi =` RHS.