Verified answer

let sin^-1(1/√ 3) = A

sin A = 1 / √ 3

cos A = √(1 - sin^2(A))

= √(1 - (1/3))

= √(2/3)

let tan^-1(-4/3) = B

tan B = -4/3

opposite side/ adj. side = 4/3===> opp. side = 4 and adj. side = 3

hypotenuse = √(4^2 + 3^2) = 5

sin B = 4/5

cos B = - 3/5

now cos[ sin^-1(1/√ 3)+tan^-1(-4/3) ]

= cos (A + B)

=> cos A cos B - sin A sin B

=> √(2/3)* (-3/5) - (1/3)(4/5)

= - (1/5)√6 - 4/15

= - (1/15) [ 4 + 3√6 ]

Calculating values for the trigonometric functions at specific angles is something like calculating the value for PI. There are several ways to estimate the value but no straightforward way to calculate it exactly. People who use trigonometry are often mostly interested in a few standard angles. (such as 30, 45, 60, and 90 degrees). For example, the sin of 30 degrees is 1/2, the cos is the square root of 3 all over 2.