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प्रश्न
Solve: `tan^-1x = cos^-1 (1 - "a"^2)/(1 + "a"^2) - cos^-1 (1 - "b"^2)/(1 + "b"^2), "a" > 0, "b" > 0`
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उत्तर
`2tan^-1x = cos^-1((1 - "a"^2)/(1 + "a"^2)) - cos^-1((1 - "b"^2)/(1 + "b"^2)) "a" > 0`
W.K.T `2tan^-1x = cos^-1((1 - x^2)/(1 + x^2))`
`cos^-1 [(1 - "a"^2)/(1 + "a"^2)] = 2tan^-1"a"`
`cos^-1[(1 - "b"^2)/(1 + "b"^2)] = 2tan^-1"b"`
R.H.S `cos^-1 [(1 - "a"^2)/(1 + "a"^2)] - cos^-1 [(1 - "b"^2)/(1 + "b"^2)]`
= `2tan^-1"a" - 2tam^-1"b"`
= `2[tan^-1"a" - tan^-1"b"]`
= `2[tan^-1 ("a" - "b")/(1 + "ab")]`
L.H.S = R.H.S
`2tan^1 [("a" - "b")/(1 + "ab")] = 2tan^-x`
x = `("a" - "b")/(1 + "ab"), "a" > 0`
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