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प्रश्न
Find the value of the following:
`tan 1/2 [sin^(-1) (2x)/(1 + x^2) + cos^(-1) (1 - y^2)/(1 + y^2)], |x| < 1, y > 0 and xy < 1`
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उत्तर
Let x = tan θ.
Then, θ = tan−1 x.
∴ `sin^(-1) (2x)/(1+x^2 ) `
= `sin^(-1) ((2tan θ)/(1 + tan^2 θ)) `
= `sin^(-1) (sin 2 θ)`
= 2θ
= 2 tan−1 x
Let y = tan `phi`.
Then, `phi` = tan−1 y.
∴ `cos^(-1) (1 - y^2)/(1+ y^2)`
= `cos^(-1) ((1 - tan^2 phi)/(1+tan^2 phi))`
= `cos^(-1)(cos 2phi)`
= `2phi`
= 2 tan−1 y
∴ `tan 1/2 [sin^(-1) (2x)/(1+ x^2) + cos^(-1) (1-y^2)/(1+y^2)]`
= `tan 1/2 [2tan^(-1) x + 2tan^(-1) y]`
= `tan[tan^(-1) x + tan^(-1) y]`
= `tan[tan^(-1) ((x+y)/(1-xy))]`
= `(x+y)/(1-xy)`
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