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Question
If `sin^-1 (2a)/(1+a^2)-cos^-1 (1-b^2)/(1+b^2)=tan^-1 (2x)/(1-x^2)`, then prove that `x=(a-b)/(1+ab)`
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Solution
Let: a = tan m
b = tan n
x = tan y
Now,
`sin^-1 (2a)/(1+a^2)-cos^-1 (1-b^2)/(1+b^2)=tan^-1 (2x)/(1-x^2)`
`=>sin^-1 (2tanm)/(1+tan^2m)-cos^-1 (1-tan^2n)/(1+tan^2n)=tan^-1 (2tany)/(1-tan^2y)`
`=>sin^-1(sin2m)-cos^-1(cos2n)=tan^-1(tan2y)` `[becausesin2x=(2tanx)/(1+tan^2x)andcos2x=(1-tan^2x)/(1+tan^2x)]`
`=>2m-2n=2y`
`=>m-n=y`
`=>tan^-1a-tan^-1b=tan^-1x` `[becausea=tanm,b=tannandx=tany]`
`=>tan^-1 (a-b)/(1+ab)=tan^-1x` `[becausetan^-1x-tan^-1y=tan^-1 (x-y)/(1+xy)]``=>(a-b)/(1+ab)=x`
`therefore(a-b)/(1+ab)=x`
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