The trick is to evaluate sin(cot^-1(5))+5sin(tan^-1(2))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)cot−1(5)
Let theta=cot^-1(5)θ=cot−1(5)
Then cot(theta) = 5cot(θ)=5
cot(theta) = "adjancent"/"opposite"cot(θ)=adjancentopposite
If I have to show that in a right triangle.
![enter image source here]()
The hypotenuse would be sqrt(1^2+5^2)√12+52
Hypotenuse = sqrt(26)=√26
sin(theta) = "Opposite"/"hypotenuse"sin(θ)=Oppositehypotenuse
sin(theta) = 1/sqrt(26)sin(θ)=1√26
Rationalizing the denominator we get
sin(theta) = (sqrt(26))/26sin(θ)=√2626
Remember thetaθ was cot^-1(5))cot−1(5))
sin(cot^-1(5)) = (sqrt(26))/26sin(cot−1(5))=√2626
Now let us work on the second part.
5sin(tan^-1(2))5sin(tan−1(2))
Theta=tan^-1(2)
tan(Theta) = 2
tan(Theta) = "Opposite"/"Adjacent"
Let us put that in a right triangle.
![enter image source here]()
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
