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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]
उत्तर
LHS = `tan^-1 ((sqrt(1 + x) - sqrt(1 - x))/(sqrt(1 + x) + sqrt(1 - x)))`
Put x = cos θ
∴ θ = cos–1x
∴ LHS = `tan^-1 ((sqrt(1 + cos theta) - sqrt(1 - cos theta))/(sqrt(1 + cos theta) + sqrt(1 - cos theta)))`
= `tan^-1 [(sqrt(2 cos^2(theta/2)) - sqrt(2 sin^2 (theta/2)))/(sqrt(2 cos^2 (theta/2)) + sqrt(2 sin^2 (theta/2)))]`
= `tan^-1 [(sqrt(2) cos (theta/2) - sqrt(2) sin (theta/2))/(sqrt(2) cos (theta/2) + sqrt(2) sin (theta/2))]`
= `tan^-1 [((sqrt(2) cos (theta/2))/(sqrt(2) cos (theta/2)) - (sqrt(2) sin (theta/2))/(sqrt(2) cos (theta/2)))/((sqrt(2) cos (theta/2))/(sqrt(2) cos (theta/2)) + (sqrt(2) sin (theta/2))/(sqrt(2) cos (theta/2)))]`
= `tan^-1 [(1 - tan(theta/2))/(1 + tan (theta/2))]`
= `tan^-1 [(tan pi/4 - tan (theta/2))/(1 + tan pi/4. tan (theta/2))]` .....`[∵ tan pi/4 =1]`
= `tan^-1 [tan (pi/4 - theta/2)]`
= `pi/4 - theta/2`
= `pi/4 - 1/2 cos^-1`x .....[∵ θ = cos–1x]
= RHS.
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