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प्रश्न
Prove the following:
`cos^-1 "x" = pi + tan^-1 (sqrt(1 - "x"^2)/"x")`, if x < 0
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उत्तर
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
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