Consider the equation . Choose the CORRECT option(s).

Asked by harshsharma2531 | 9th Aug, 2020, 09:28: AM

Expert Answer:

To find solutions of the equation tan-1(x-1/x-2) + tan-1(x+1/x+2) = pi/4
tan to the power of negative 1 end exponent open parentheses fraction numerator straight x minus 1 over denominator straight x minus 2 end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses fraction numerator straight x plus 1 over denominator straight x plus 2 end fraction close parentheses equals straight pi over 4
rightwards double arrow tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style fraction numerator straight x minus 1 over denominator straight x minus 2 end fraction plus fraction numerator straight x plus 1 over denominator straight x plus 2 end fraction end style over denominator 1 minus begin display style fraction numerator straight x minus 1 over denominator straight x minus 2 end fraction end style cross times begin display style fraction numerator straight x plus 1 over denominator straight x plus 2 end fraction end style end fraction close parentheses equals straight pi over 4
rightwards double arrow tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style open parentheses straight x minus 1 close parentheses open parentheses straight x plus 2 close parentheses plus open parentheses straight x plus 1 close parentheses open parentheses straight x minus 2 close parentheses end style over denominator straight x squared minus 4 minus open parentheses straight x squared minus 1 close parentheses end fraction close parentheses equals straight pi over 4
rightwards double arrow tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style straight x squared plus straight x minus 2 plus straight x squared minus straight x minus 2 end style over denominator straight x squared minus 4 minus open parentheses straight x squared minus 1 close parentheses end fraction close parentheses equals straight pi over 4
rightwards double arrow tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style 2 straight x squared minus 4 end style over denominator negative 3 end fraction close parentheses equals straight pi over 4
rightwards double arrow fraction numerator begin display style 2 straight x squared minus 4 end style over denominator negative 3 end fraction equals tan straight pi over 4 equals 1
rightwards double arrow fraction numerator 2 straight x squared minus 4 over denominator negative 3 end fraction equals 1
rightwards double arrow 2 straight x squared minus 4 equals negative 3
rightwards double arrow 2 straight x squared equals 1
rightwards double arrow straight x equals plus-or-minus fraction numerator 1 over denominator square root of 2 end fraction
Sum space of space all space solutions equals fraction numerator 1 over denominator square root of 2 end fraction minus fraction numerator 1 over denominator square root of 2 end fraction equals 0
Number space of space solutions space equals space 2
Sum space of space squares space of space all space solutions equals 1 half plus 1 half equals 1

Answered by Renu Varma | 9th Aug, 2020, 04:32: PM