Question

cotan(cos-1(3/5))+tan(sin-1(3/5))

Answer 1

seems your question is -----> $cot[cos^{-1}(3/5)]+tan[sin^{-1}(3/5)]$

cos^-1(3/5) = A => cosA = 3/5
cotA = 3/4 => A = cot^-1(3/4)
hence, you can write cos^-1(3/5) = cot^-1(3/4)

similarly, sin^-1(3/5) = B => sinB = 3/5
tanB = 3/4 => tan^-1(3/4) = B
hence, you can also write sin^-1(3/5) = tan^-1(3/4)

now, cot{cot^-1(3/4)} + tan{tan^-1(3/4)}

we know, tan(tan^-1x) = x
cot(cot^-1x) = x

so,cot{cot^-1(3/4)} = 3/4
tan{tan^-1(3/4)} = 3/4

hence, 3/4 + 3/4 = 3/2