`tan^(-1)(sinx)=sin^(-1)(tanx)` holds true for

Options

A. `"x" epsilonR`
B. `2npi-(pi)/2lexle2npi+(pi)/2("n" epsilonz)`
C. `"x" epsilon{0,z^(+)}`
D. `"x" epsilonnpi("n" epsilonz)`

Answer

D

Explanation

`tan^(-1)(sinx)=sin^(-1)(tanx)`
`impliestan^(-1)(sinx)=tan^(-1)((tanx)/(sqrt(1-tan^(2)x)))`
`impliessinx=(tanx)/(sqrt(1-tan^(2)x))`
`implies sinx=0` or `cos^(2)x(1-tan^(2)x)x=1`
`impliessinx=0` or `cos2x=1`
`impliesx=npi`