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Inverse Trigonometric Functions

Question

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Prove that $tan to the power of negative 1 end exponent 1 over 7 plus tan to the power of negative 1 end exponent 2 over 9 equals 1 half cos to the power of negative 1 end exponent 3 over 5$

Solution

straight L. straight H. straight S. equals tan to the power of negative 1 end exponent 1 over 7 plus tan to the power of negative 1 end exponent 1 over 13 equals tan to the power of negative 1 end exponent open square brackets fraction numerator begin display style 1 over 7 end style plus begin display style 1 over 13 end style over denominator 1 minus begin display style 1 over 7 end style cross times begin display style 1 third end style end fraction close square brackets space space space 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 begin display style fraction numerator 13 plus 7 over denominator 91 end fraction end style over denominator begin display style fraction numerator 91 minus 1 over denominator 91 end fraction end style end fraction close square brackets equals tan to the power of negative 1 end exponent open parentheses 20 over 90 close parentheses equals tan to the power of negative 1 end exponent open parentheses 2 over 9 close parentheses equals straight R. straight H. straight S.

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