If a>b>c>0 then prove

cot¯¹(ab+1/a-b)+cot¯¹(bc+1/b-c)+cot¯¹(ca+1/c-a)=¶

Asked by patelmanan4102002 | 5th Aug, 2019, 08:21: PM

Expert Answer:

cot to the power of negative 1 end exponent open parentheses fraction numerator ab plus 1 over denominator straight a minus straight b end fraction close parentheses plus cot to the power of negative 1 end exponent open parentheses fraction numerator bc plus 1 over denominator straight b minus straight c end fraction close parentheses plus cot to the power of negative 1 end exponent open parentheses fraction numerator ca plus 1 over denominator straight c minus straight a end fraction close parentheses
equals tan to the power of negative 1 end exponent open parentheses fraction numerator straight a minus straight b over denominator 1 plus ab end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses fraction numerator straight b minus straight c over denominator 1 plus bc end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses fraction numerator straight c minus straight a over denominator 1 plus ca end fraction close parentheses
equals tan to the power of negative 1 end exponent straight a minus tan to the power of negative 1 end exponent straight b plus tan to the power of negative 1 end exponent straight b minus tan to the power of negative 1 end exponent straight c plus tan to the power of negative 1 end exponent straight c minus tan to the power of negative 1 end exponent straight a

As space straight a greater than straight b greater than straight c space rightwards double arrow straight a minus straight b greater than 0 comma space straight b minus straight c greater than 0 space but space straight c minus straight a less than 0 space and space since space the space period space of space tan space is space straight pi
Therefore comma space tan to the power of negative 1 end exponent open parentheses fraction numerator straight a minus straight b over denominator 1 plus ab end fraction close parentheses equals tan to the power of negative 1 end exponent straight a minus tan to the power of negative 1 end exponent straight b comma space tan to the power of negative 1 end exponent open parentheses fraction numerator straight b minus straight c over denominator 1 plus bc end fraction close parentheses equals tan to the power of negative 1 end exponent straight b minus tan to the power of negative 1 end exponent straight c space & space space tan to the power of negative 1 end exponent open parentheses fraction numerator straight c minus straight a over denominator 1 plus ca end fraction close parentheses space equals tan to the power of negative 1 end exponent straight c minus tan to the power of negative 1 end exponent straight a plus straight pi
rightwards double arrow tan to the power of negative 1 end exponent open parentheses fraction numerator straight a minus straight b over denominator 1 plus ab end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses fraction numerator straight b minus straight c over denominator 1 plus bc end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses fraction numerator straight c minus straight a over denominator 1 plus ca end fraction close parentheses equals space tan to the power of negative 1 end exponent straight a minus tan to the power of negative 1 end exponent straight b plus tan to the power of negative 1 end exponent straight b minus tan to the power of negative 1 end exponent straight c plus tan to the power of negative 1 end exponent straight c minus tan to the power of negative 1 end exponent straight a plus straight pi equals straight pi
Hence comma space cot to the power of negative 1 end exponent open parentheses fraction numerator ab plus 1 over denominator straight a minus straight b end fraction close parentheses plus cot to the power of negative 1 end exponent open parentheses fraction numerator bc plus 1 over denominator straight b minus straight c end fraction close parentheses plus cot to the power of negative 1 end exponent open parentheses fraction numerator ca plus 1 over denominator straight c minus straight a end fraction close parentheses equals straight pi

Answered by Renu Varma | 6th Aug, 2019, 10:26: AM