4(cot-1(3)+cosec-1(5))=4(tan-1(1/3)+sin-1(1/√5)).
Let x=tan-1(1/3) and y=sin-1(1/√5), then:
tan(x)=1/3, so sin(x)=1/√10, cos(x)=3/√10, and sin(y)=1/√5, cos(y)=2/√5
sin(x+y)=sin(x)cos(y)+cos(x)sin(y)=
2/√50+3/√50=5/√50=√50/10=5√2/10=√2/2=sin(π/4).
Therefore sin(x+y)=sin(tan-1(1/3)+sin-1(1/√5))=sin(π/4).
So, tan-1(1/3)+sin-1(1/√5)=π/4, and 4(tan-1(1/3)+sin-1(1/√5))=π QED