What does #sin(arc cot(5))+5sin(arc tan(2))# equal?

The trick is to evaluate #sin(cot^-1(5))+5sin(tan^-1(2))# is
breaking down the problem and solving each part separately.

Let us start with evaluating #cot^-1(5)#

Let #theta=cot^-1(5)#

Then #cot(theta) = 5#

#cot(theta) = "adjancent"/"opposite"#

If I have to show that in a right triangle.

The hypotenuse would be #sqrt(1^2+5^2)#
Hypotenuse #= sqrt(26)#

#sin(theta) = "Opposite"/"hypotenuse"#
#sin(theta) = 1/sqrt(26)#

Rationalizing the denominator we get
#sin(theta) = (sqrt(26))/26#

Remember #theta# was #cot^-1(5))#

#sin(cot^-1(5)) = (sqrt(26))/26#

Now let us work on the second part.

#5sin(tan^-1(2))#
#Theta=tan^-1(2)#
#tan(Theta) = 2#
#tan(Theta) = "Opposite"/"Adjacent"#

Let us put that in a right triangle.

Now to find the hypotenuse

Hypotenuse #=sqrt(2^2+1^2)#
Hypotenuse #=sqrt(5)#

#sin(Theta) = "Opposite"/"hypotenuse"#

#sin(Theta) =2/sqrt(5)#

Rationalizing the denominator

#sin(Theta)=(2sqrt(5))/5#

Remember #Theta=tan^-1(2)#

#sin(tan^-1(2))=(2sqrt(5))/5#

#5sin(tan^-1(2))=5**(2sqrt(5))/5#

#5sin(tan^-1(2))=2sqrt(5)#

Now the final steps

#sin(cot^-1(5))+5sin(tan^-1(2)) = sqrt(26)/26+2sqrt(5)#

#sin(cot^-1(5))+5sin(tan^-1(2)) = (sqrt(26)+52sqrt(5))/26#