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प्रश्न
Prove that `tan^-1 ((sqrt(1 + x^2) + sqrt(1 - x^2))/((1 + x^2) - sqrt(1 - x^2))) = pi/2 + 1/2 cos^-1x^2`
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उत्तर
L.H.S. `tan^-1 [(sqrt(1 + x^2) + sqrt(1 - x^2))/(sqrt(1 + x^2 - sqrt(1 - x^2)))]`
Put x2 = cos θ
∴ θ = `cos^-1 x^2`
⇒ `tan^-1 [(sqrt(1 + cos theta) + sqrt(1 - cos theta))/(sqrt(1 + cos theta) - sqrt(1 - cos theta))]`
⇒ `tan^-1 [sqrt(2cos^2 theta/2 + sqrt(2sin^2 theta/2))/(sqrt(2cos^2 theta/2 - sqrt(2sin^2 theta/2)))]` ......`[(because 1 + cos theta = 2 cos^2 theta/2),(1 - cos theta = 2 sin^2 theta/2)]`
⇒ `tan^-1 [(cos theta/2 + sin theta/2),(cos theta/2 - sin theta/2)]`
⇒ `tan^-1 [(1 + tan theta/2),(1 - tan theta/2)]` ......[Dividing the Nr. and Den. by cos θ/2]
⇒ `tan [tan(pi/4 theta/2)]` ......`[because (1 + tan theta)/(1 - tan theta) = tan(pi/4 + theta)]`
⇒ `pi/4 + theta/2`
⇒ `pi/4 + 1/2 cos^-1 x^2` R.H.S. ......[Putting θ = cos–1x2]
Hence proved.
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