Question

cos[(tan^-1(3/4))+(cos^-1(9/41))]

Answer 1

let tan^-1 (3/4) = A and let cos^-1 (9/41) = B

then cos[(tan^-1(3/4))+(cos^-1(9/41))]
= cos(A + B)
= cosA cosB - sinA sinB

IF tan^-1 (3/4) = A
then tanA = 3/4
---> sinA = 3/5 and cosA = 4/5, recognize the 3-4-5 right angled triangle

if cos^-1 (9/41) = B
then cosB = 9/41
---> x^2 + y^2 = r^2
81 + y^2 = 1681
y^2 = 1600
y = √1600 = 40 and thus sinB = 40/41

so back to
cosA cosB - sinA sinB
= (4/5)(9/41) - (3/5)(40/41) = -84/205

( I tested my answer with my calculator, it is correct)