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If x >1 , then 2\ tan^(-1)x+sin^(-1)((2x...

If `x >1` , then `2\ tan^(-1)x+sin^(-1)((2x)/(1+x^2))` is equal to `4tan^(-1)x` (b) 0 (c) `pi/2` (d) `pi`

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We have to find
`2\ tan^(-1)x+sin^(-1)((2x)/(1+x^2))`
For `x>1`
`2\ tan^(-1)x=pi-sin^(-1)((2x)/(1+x^2))`
Therefore,
`2\ tan^(-1)x+sin^(-1)((2x)/(1+x^2))`
`=pi-sin^(-1)((2x)/(1+x^2))+sin^(-1)((2x)/(1+x^2))`
`=pi`
...
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