Question

Prove the following:

`cos^-1 "x" = pi + tan^-1 (sqrt(1 - "x"^2)/"x")`, if x < 0

Answer 1

Let cos^-1 x = α

Then, cos α = x, where 0 < α < π

Since, x < 0, `pi/2` < α < π

Now, `tan^-1 (sqrt(1 - "x"^2)/"x") = tan^-1  (sqrt(1 - cos^2 alpha)/cos alpha)`

= tan^-1 (tan α)    ....(1)

But `pi/2 < alpha < pi`, therefore inverse of tangent does not exist.

Consider, `pi/2 - pi < alpha - pi < pi - pi`,

`therefore - pi/2 < alpha - pi < 0` 

and tan (α - π) = tan [ - (π - α)]
= - tan (π - α)      .....[∵ tan (- θ) = - tan θ]

= - (- tan α) = tan α

∴ from (1), we get

`tan^-1 (sqrt(1 - "x"^2)/"x") = tan^-1 [tan (alpha - pi)]`

`= alpha - pi       ......[∵ tan^-1 (tan"x") = "x"]`

`= cos^-1 "x" - pi`

∴ `cos^-1 "x" = pi + tan^-1 (sqrt(1 - "x"^2)/"x")`, if x < 0