Solve begin mathsize 20px style tan to the power of negative 1 end exponent open parentheses 2 x close parentheses plus tan to the power of negative 1 end exponent open parentheses 3 x close parentheses equals bevelled straight pi over 4 end style

Asked by sunil2791 | 21st Apr, 2017, 07:43: PM

Expert Answer:

begin mathsize 16px style We space know space that comma
tan to the power of negative 1 end exponent straight x plus tan to the power of negative 1 end exponent straight y space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator straight x plus straight y over denominator 1 minus xy end fraction close parentheses
Using space the space above space we space get comma
tan to the power of negative 1 end exponent 2 straight x plus tan to the power of negative 1 end exponent 3 straight x space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x plus 3 straight x over denominator 1 minus left parenthesis 2 straight x right parenthesis left parenthesis 3 straight x right parenthesis end fraction close parentheses
rightwards double arrow tan to the power of negative 1 end exponent 2 straight x plus tan to the power of negative 1 end exponent 3 straight x space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 5 straight x over denominator 1 minus 6 straight x squared end fraction close parentheses
Given space that space tan to the power of negative 1 end exponent 2 straight x plus tan to the power of negative 1 end exponent 3 straight x space equals straight pi over 4
rightwards double arrow tan to the power of negative 1 end exponent open parentheses fraction numerator 5 straight x over denominator 1 minus 6 straight x squared end fraction close parentheses equals straight pi over 4
rightwards double arrow fraction numerator 5 straight x over denominator 1 minus 6 straight x squared end fraction equals tan straight pi over 4
rightwards double arrow fraction numerator 5 straight x over denominator 1 minus 6 straight x squared end fraction equals 1
rightwards double arrow 5 straight x equals 1 minus 6 straight x squared
rightwards double arrow 6 straight x squared plus 5 straight x minus 1 equals 0
rightwards double arrow left parenthesis 6 straight x minus 1 right parenthesis left parenthesis straight x plus 1 right parenthesis equals 0
rightwards double arrow straight x equals 1 over 6 space space or space space straight x equals negative 1
end style

Answered by Rebecca Fernandes | 27th Nov, 2017, 01:16: PM