Inverse Trigonometric Functions

Question

Prove that $2 space tan to the power of negative 1 end exponent 1 half plus tan to the power of negative 1 end exponent 1 over 7 equals tan to the power of negative 1 end exponent 31 over 17.$

Solution

$L italic. H italic. S italic. italic space italic space 2 space tan to the power of negative 1 end exponent 1 half plus tan to the power of negative 1 end exponent 1 over 7 space space space space space space space space space space space equals tan to the power of negative 1 end exponent open square brackets fraction numerator 2 cross times begin display style 1 half end style over denominator 1 minus open parentheses begin display style 1 half end style close parentheses squared end fraction close square brackets plus tan to the power of negative 1 end exponent 1 over 7 space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space 2 space tan to the power of negative 1 end exponent space straight x space equals space tan to the power of negative 1 end exponent space fraction numerator 2 straight x over denominator 1 minus straight x squared end fraction close square brackets space space space space space space space space space space space equals space tan to the power of negative 1 end exponent fraction numerator 1 over denominator 1 minus begin display style 1 fourth end style end fraction plus tan to the power of negative 1 end exponent 1 over 7 equals tan to the power of negative 1 end exponent 4 over 3 plus tan to the power of negative 1 end exponent 1 over 7 space space space space space space space space space space space equals space tan to the power of negative 1 end exponent open square brackets fraction numerator begin display style 4 over 3 plus 1 over 7 end style over denominator 1 minus begin display style 4 over 3 end style cross times begin display style 1 over 7 end style end fraction close square brackets equals space tan to the power of negative 1 end exponent open square brackets fraction numerator 28 plus 3 over denominator 21 minus 4 end fraction close square brackets equals tan to the power of negative 1 end exponent 31 over 17 space space space space space space space space space space space space equals space straight R. space straight H. space straight S.$

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Inverse Trigonometric Functions

Prove that $2 space tan to the power of negative 1 end exponent 1 half plus tan to the power of negative 1 end exponent 1 over 7 equals tan to the power of negative 1 end exponent 31 over 17.$

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