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Question
Find the value of 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`
Solution
Let x = tan θ. Then, θ = tan−1 x.
`:. sin^(-1) (2x)/(1+x^2 ) `
`= sin^(-1) ((2tan theta)/(1 + tan^2 theta)) `
`= sin^(-1) (sin 2 theta)`
` = 2theta `
`= 2 tan^(-1) x`
Let y = tan Φ. Then, Φ = 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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