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`tan^(-1)(sinx)=sin^(-1)(tanx)` holds true forA. `"x" epsilonR`B. `2npi-(pi)/2lexle2npi+(pi)/2("n" epsilonz)`C. `"x" epsilon{0,z^(+)}`D. `"x" epsilonnpi("n" epsilonz)` |
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Answer» Correct Answer - D `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` |
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