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Inverse Trigonometric Functions

Question
CBSEENMA12032828
Wired Faculty App

sin (tan–1 x), | x | < 1 is equal to
  • fraction numerator straight x over denominator square root of 1 minus straight x squared end root end fraction
  • fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction
  • fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction
  • fraction numerator straight x over denominator square root of 1 minus straight x squared end root end fraction

Solution

D.

fraction numerator straight x over denominator square root of 1 minus straight x squared end root end fraction Put tan–1 x = θ, | x | < 1

therefore space space space space space space space straight x equals tan space straight theta comma space space open vertical bar straight theta close vertical bar less than straight pi over 4 therefore space space space space space space sin left parenthesis tan to the power of negative 1 end exponent space straight x right parenthesis equals sin space straight theta space equals fraction numerator sin space straight theta over denominator cos space straight theta end fraction cos space straight theta equals tan space straight theta space cos space straight theta space equals fraction numerator tan space straight theta over denominator sec space straight theta end fraction space space space space space space space space equals fraction numerator tsn space straight theta over denominator square root of 1 plus tan squared space straight theta end root end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket because space sec space straight theta space greater than space 0 space abd space open vertical bar straight theta close vertical bar space less than straight pi over 4 right square bracket space therefore space space space sin space left parenthesis tan to the power of negative 1 end exponent space straight x right parenthesis space equals space fraction numerator straight x over denominator square root of 1 plus straight x squared end root end fraction therefore space space space space left parenthesis straight D right parenthesis space is space correct space answer.

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