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Inverse Trigonometric Functions

Question
CBSEENMA12032782
Wired Faculty App

Prove 2 space tan to the power of negative 1 end exponent space straight x space equals space sin to the power of negative 1 end exponent fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction equals tan to the power of negative 1 end exponent fraction numerator 2 straight x over denominator 1 minus straight x squared end fraction

Solution

Put x = tan straight theta
Now space 2 space tan to the power of negative 1 end exponent space straight x equals 2 space tan to the power of negative 1 end exponent space straight x equals 2 space tan to the power of negative 1 end exponent left parenthesis tan space straight theta right parenthesis equals 2 space straight theta therefore space space space 2 space tan to the power of negative 1 end exponent space straight x equals 2 space straight theta 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 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 parenthesis 1 right parenthesis Also space space space sin to the power of negative 1 end exponent space fraction numerator 2 space straight x over denominator 1 plus straight x squared end fraction equals sin to the power of negative 1 end exponent open parentheses fraction numerator 2 space tan space straight theta over denominator 1 plus tan space straight theta end fraction close parentheses equals sin to the power of negative 1 end exponent left parenthesis sin space 2 space straight theta right parenthesis equals 2 space straight theta therefore space space space space space space sin to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction close parentheses equals 2 space straight theta 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 space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis Again space tan to the power of negative 1 end exponent space fraction numerator 2 space straight x over denominator 1 minus straight x squared end fraction equals tan to the power of negative 1 end exponent open parentheses fraction numerator 2 space tan space straight theta over denominator 1 minus tan space straight theta end fraction close parentheses equals sin to the power of negative 1 end exponent left parenthesis sin space 2 space straight theta right parenthesis therefore space space space space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 space straight x over denominator 1 minus straight x squared end fraction close parentheses equals 2 space straight theta 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 space space space space space space space space space space space space space space space space space.... left parenthesis 3 right parenthesis
From (1), (2), (3) we get,
2 space tan to the power of negative 1 end exponent space straight x equals sin to the power of negative 1 end exponent space open parentheses fraction numerator 2 space straight x over denominator 1 plus straight x squared end fraction close parentheses equals tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x over denominator 1 minus straight x squared end fraction close parentheses

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