Question

Question #2cfd2

Answer 1

`-sqrt3.`

Explanation:

By the Defn. of `cos^-1` function,

`cos^-1 x=theta, |x| le 1 iff costheta=x, theta in [0,pi].`

Since, `cos (pi-pi/3)=cos(2pi/3)=-cos(pi/3)=-1/2,`

`&, 2pi/3 in [0,pi],` we have, by Defn.,

`cos^-1(-1/2)=2pi/3.`

`rArr tan{cos^-1(-1/2)}=tan(2pi/3)=tan(pi-pi/3)=-tan(pi/3)=-sqrt3.`