Question

$Prove space that space tan to the power of negative 1 end exponent 1 fifth plus tan to the power of negative 1 end exponent 1 over 7 plus tan to the power of negative 1 end exponent 1 third plus tan to the power of negative 1 end exponent 1 over 8 equals straight pi over 4.$

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Solution

straight L. straight H. straight S. equals open parentheses tan to the power of negative 1 end exponent 1 fifth plus tan to the power of negative 1 end exponent 1 over 7 close parentheses plus open parentheses tan to the power of negative 1 end exponent 1 third plus tan to the power of negative 1 end exponent 1 over 8 close parentheses equals space tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style 1 fifth end style plus begin display style 1 over 7 end style over denominator 1 minus begin display style 1 fifth end style space begin display style 1 over 7 end style end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style 1 third end style plus begin display style 1 over 8 end style over denominator 1 minus begin display style 1 third end style space begin display style 1 over 8 end style end fraction close parentheses equals tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style fraction numerator 7 plus 5 over denominator 35 end fraction end style over denominator begin display style fraction numerator 35 minus 1 over denominator 24 end fraction end style end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style fraction numerator 8 plus 3 over denominator 24 end fraction end style over denominator begin display style fraction numerator 24 minus 1 over denominator 24 end fraction end style end fraction close parentheses equals space tan to the power of negative 1 end exponent open parentheses 12 over 24 close parentheses plus tan to the power of negative 1 end exponent open parentheses 11 over 23 close parentheses equals tan to the power of negative 1 end exponent open parentheses 6 over 17 close parentheses plus tan to the power of negative 1 end exponent open parentheses 11 over 23 close parentheses equals space tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style 6 over 17 end style plus begin display style 11 over 13 end style over denominator 1 minus begin display style 6 over 17 end style cross times begin display style 11 over 23 end style end fraction close parentheses equals tan to the power of negative 1 end exponent open parentheses 6 over 17 close parentheses plus tan to the power of negative 1 end exponent open parentheses 11 over 23 close parentheses equals space tan to the power of negative 1 end exponent open parentheses 325 over 325 close parentheses equals tan to the power of negative 1 end exponent space 1 equals straight pi over 4 equals straight R. straight H. straight S.

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Inverse Trigonometric Functions

$Prove space that space tan to the power of negative 1 end exponent 1 fifth plus tan to the power of negative 1 end exponent 1 over 7 plus tan to the power of negative 1 end exponent 1 third plus tan to the power of negative 1 end exponent 1 over 8 equals straight pi over 4.$

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