`sin^(-1)(3x)/5+sin^(-1)(4x)/5=sin^(-1)x`, then roots of theequation are-a.`0`b. `1`c.`-1 `d. `-2`

`sin^(-1)(3x)/5+sin^(-1)(4x)/5=sin^(-1)x`, then roots of theequation a

1.	`sin^(-1)(3x)/5+sin^(-1)(4x)/5=sin^(-1)x`, then roots of theequation are-a.`0`b. `1`c.`-1 `d. `-2`
Answer» Correct Answer - D `"sin"^(-1)(3x)/(5)+"sin"^(-1)(4x)/(5)=sin^(-1)x` `rArr sin^(-1)((3x)/(5)sqrt(1-(16x^(2))/(25))+(4x)/(5)sqrt(1-(9x^(2))/(25)))=sin^(-1)x` `rarr (3x)/(5)(sqrt(25-16x^(2)))/(5)+(4x)/(5)(sqrt(25-9x^(2)))/(5)=x` `rArr x = 0` or `3sqrt(25-16x^(2))+4sqrt(25-9x^(2))=25` Now `sqrt(225-144x^(2))+sqrt(400-144x^(2))=25` .....(1) `rArr (175)/(sqrt(400-144x^(2))-sqrt(225-144x^(2)))=25` `rArr sqrt(400-144x^(2))-sqrt(225-144x^(2))=7` ....(2) From (1) and (2), `sqrt(400-144x^(2))=16` `rArr 400-144x^(2)=256` `rArr x^(2)=1` `rArr x = pm 1`

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