Question

The solution set of the equation `sin^(-1)x=2 tan^(-1)x` is
A. `{1,2}`
B. `{-1,2}`
C. `{-1,1}`
D. `{1,1//2,0}`

Answer 1

Clearly LHS of the given equation is meaningful for `x in [-1,1]` and RHS is defined for all x`x in R` So the given equation exists for `x in [-1,1]`
Now
`sin^(-1)x=2 tan^(-1)`
`rarr sin^(-1) x= sin^(-1)((2x)/(1+x^(2)))`
`rarr x^(3)=x rarr x=0,1,-1`