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प्रश्न
Prove that:
`tan^-1 ((sqrt(1 + x) - sqrt(1 - x))/(sqrt(1 + x) + sqrt(1 - x))) = pi/4 - 1/2 cos^-1 x`, for `- 1/sqrt2 ≤ x ≤ 1`
[Hint: Put x = cos 2θ]
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उत्तर
Put x = cos θ
∴ θ = cos–1 x
L.H.S. = `tan^-1 ((sqrt(1 + x) - sqrt(1 - x))/(sqrt(1 + x) + sqrt(1 - x)))`
= `tan^-1 ((sqrt(1 + cos θ) - sqrt(1 - cos θ))/(sqrt(1 + cos θ) + sqrt(1 - cos θ)))`
= `tan^-1 [(sqrt(2 cos^2(θ/2)) - sqrt(2 sin^2 (θ/2)))/(sqrt(2 cos^2 (θ/2)) + sqrt(2 sin^2 (θ/2)))]`
= `tan^-1 [(sqrt(2) cos (θ/2) - sqrt(2) sin (θ/2))/(sqrt(2) cos (θ/2) + sqrt(2) sin (θ/2))]`
= `tan^-1 [((sqrt(2) cos (θ/2))/(sqrt(2) cos (θ/2)) - (sqrt(2) sin (θ/2))/(sqrt(2) cos (θ/2)))/((sqrt(2) cos (θ/2))/(sqrt(2) cos (θ/2)) + (sqrt(2) sin (θ/2))/(sqrt(2) cos (θ/2)))]`
= `tan^-1 [(1 - tan(θ/2))/(1 + tan (θ/2))]`
= `tan^-1 [(tan pi/4 - tan (θ/2))/(1 + tan pi/4. tan (θ/2))] ....[∵ tan pi/4 =1]`
= `tan^-1 [tan (pi/4 - θ/2)]`
= `pi/4 - θ/2`
= `pi/4 - 1/2 cos^-1`x .....[∵ θ = cos–1 x]
∴ L.H.S. = R.H.S.
