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Inverse Trigonometric Functions

Question

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Simplify

$tan to the power of negative 1 end exponent open parentheses fraction numerator square root of 1 plus x squared end root minus 1 over denominator x end fraction close parentheses$

Solution

Let space straight z equals tan to the power of negative 1 end exponent space open parentheses fraction numerator square root of 1 plus straight x squared end root minus 1 over denominator straight x end fraction close parentheses Put space space space straight x equals tan space straight theta therefore space space space space straight z equals tan to the power of negative 1 end exponent space open parentheses fraction numerator square root of 1 plus tan squared space straight theta end root over denominator tan space straight theta end fraction close parentheses equals tan to the power of negative 1 end exponent open parentheses fraction numerator sec space straight theta minus 1 over denominator tan space straight theta end fraction close parentheses equals tan to the power of negative 1 end exponent space open square brackets fraction numerator begin display style fraction numerator 1 over denominator cos space straight theta end fraction minus 1 end style over denominator begin display style fraction numerator sin space straight theta over denominator cos space straight theta end fraction end style end fraction close square brackets space space space space space space space space space equals tan to the power of negative 1 end exponent open parentheses fraction numerator 1 minus cos space theta over denominator sin space theta end fraction close parentheses equals tan to the power of negative 1 end exponent open square brackets fraction numerator 2 space sin squared space begin display style theta over 2 end style over denominator 2 space sin space begin display style theta over 2 end style space cos space begin display style theta over 2 end style end fraction close square brackets space space space space space space space space space equals space tan to the power of negative 1 end exponent open parentheses tan space theta over 2 close parentheses equals theta over 2 equals 1 half tan to the power of negative 1 end exponent space x space space space space space space space space space

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