`2tan^(-1)(-2)` is equal to (a)`-cos^(-1)((-3)/5)` (b) `-pi+cos^(-1)3/5` (c)`-pi/2+tan^(-1)(-3/4)` (d) `-picot^(-1)(-3/4)`

Options

A. `-cos^(-1) ((-3)/(5))`
B. `-pi + cos^(-1).(3)/(5)`
C. `-(pi)/(2) + tan^(-1) (-(3)/(4))`
D. `-pi + cot^(-1) (-(3)/(4))`

Answer

A, B, C

Explanation

Let `tan^(-1) (-2) = theta " or " tan theta = -2`
`rArr theta in (-pi//2, 0) " or " 2 theta in (-pi, 0)`
`cos (-2 theta) = cos 2 theta = (1 - tan^(2) theta)/(1 + tan^(2) theta) = (-3)/(5)`
or `-2 theta = cos^(-1) ((-3)/(5)) = pi - cos^(-1).(3)/(5)`
or `2 theta = pi + cos^(-1).(3)/(5) = - pi + tan^(-1).(4)/(3)`
`= - pi + cot^(-1).(3)/(4) = - pi + (pi)/(2) - tan^(-1). (3)/(4)`
`= - (pi)/(2) - tan^(-1).(3)/(4) = - (pi)/(2) + tan^(-1) (-(3)/(4))`