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Inverse Trigonometric Functions

Question
CBSEENMA12032805
Wired Faculty App

Prove that tan to the power of negative 1 end exponent x plus tan to the power of negative 1 end exponent fraction numerator 2 space x over denominator 1 minus x squared end fraction equals tan to the power of negative 1 end exponent open parentheses fraction numerator 3 x minus x cubed over denominator 1 minus 3 x squared end fraction close parentheses equals tan to the power of negative 1 end exponent open parentheses fraction numerator 3 x minus x cubed over denominator 1 minus 3 x squared end fraction close parentheses comma open vertical bar x close vertical bar less than fraction numerator 1 over denominator square root of 3 end fraction.

Solution
Put x = tan θ
straight L. straight H. straight S. space equals space tan to the power of negative 1 end exponent space straight x plus 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 space space space space space space space space space equals space tan to the power of negative 1 end exponent left parenthesis tan space straight theta right parenthesis plus tan to the power of negative 1 end exponent open parentheses fraction numerator 2 space tan space straight theta over denominator 1 minus tan squared space straight theta end fraction close parentheses space space space space space space space space space equals space tan to the power of negative 1 end exponent left parenthesis tan space straight theta right parenthesis plus tan to the power of negative 1 end exponent 9 tan space 2 space straight theta right parenthesis equals straight theta plus 2 space straight theta equals 3 space straight theta straight R. straight H. straight S space equals tan to the power of negative 1 end exponent open parentheses fraction numerator 3 straight x minus straight x cubed over denominator 1 minus 3 straight x squared end fraction close parentheses equals tan to the power of negative 1 end exponent open parentheses fraction numerator 3 space tan space straight theta minus tan cubed space straight theta over denominator 1 minus 3 space tan squared space straight theta end fraction close parentheses space space space space space space space space space equals space tan to the power of negative 1 end exponent left parenthesis tan space 3 straight theta right parenthesis equals 3 therefore space space straight L. straight H. straight S. space equals space straight R. straight H. straight S.

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