Question

Find the value of  `sin^-1 (- 1) + cos^-1 (1/2) + cot^-1 (2)`

Answer 1

x = `sin^-1 (1)`

sin x = `-1 = sin(- pi/2)`

x = `- pi/2`

y = `cos^-1 (1/2)`

cos y = `1/2 = cos  pi/3`

y = `pi/3`

z = `cot^-1 (2)`

cot z = 2

z = `cot^-1 (2)` is constant.

`sin^-1(- 1) + cos^-1(1/3) + cot^-1(2)`

= `- pi/2 + pi/3 + cot^-1(2)`

= `- (3pi + 2pi)/6 + cot^-1(2)`

= `cot^-1(2) - pi/6`