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Inverse Trigonometric Functions

Question

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Prove that $2 space sin to the power of negative 1 end exponent 3 over 5 equals tan to the power of negative 1 end exponent 24 over 7.$

Solution

Let space sin to the power of negative 1 end exponent 3 over 5 equals straight theta. space Then space sin space straight theta equals 3 over 5 therefore space space space cos space straight theta space equals square root of 1 minus sin squared space straight theta end root equals square root of 1 minus 9 over 25 end root equals square root of 16 over 25 end root equals 4 over 5 therefore space space space tan space straight theta space equals space fraction numerator sin space straight theta over denominator cos space straight theta end fraction equals 3 over 5 cross times 5 over 4 equals 3 over 4 space space space space rightwards double arrow space space straight theta equals space tan to the power of negative 1 end exponent space 3 over 4 Now comma space space space 2 space sin to the power of negative 1 end exponent 3 over 5 equals 2 space straight theta equals 2 space tan to the power of negative 1 end exponent space 3 over 4 space space space space space space space space space space space space space space space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 cross times begin display style 3 over 4 end style over denominator 1 minus begin display style 9 over 16 end style end fraction close parentheses equals tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style 6 over 4 end style over denominator begin display style 7 over 16 end style end fraction close parentheses equals tan to the power of negative 1 end exponent open parentheses 6 over 4 cross times 16 over 7 close parentheses equals tan to the power of negative 1 end exponent 24 over 7 therefore space space space space space space 2 space sin to the power of negative 1 end exponent 3 over 5 equals tan to the power of negative 1 end exponent 24 over 7

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