$tan to the power of negative 1 end exponent open parentheses straight x over straight y close parentheses minus tan to the power of negative 1 end exponent fraction numerator straight x minus straight y over denominator straight x plus straight y end fraction is space equal space to$
$straight pi over 2$
$straight pi over 3$
$straight pi over 4$
$fraction numerator negative 3 straight pi over denominator 4 end fraction$

Solution

tan to the power of negative 1 end exponent open parentheses x over y close parentheses minus tan to the power of negative 1 end exponent open parentheses fraction numerator x minus y over denominator x plus y end fraction close parentheses equals tan to the power of negative 1 end exponent open parentheses x over y close parentheses minus tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style x over y end style minus 1 over denominator begin display style x over y end style plus 1 end fraction close parentheses space space space 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 parentheses x over y close parentheses minus open curly brackets tan to the power of negative 1 end exponent x over y minus tan to the power of negative 1 end exponent space 1 close curly brackets equals tan to the power of negative 1 end exponent space 1 equals straight pi over 4

∴ (C) is correct answer.

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