tan¹x + tan¹y=π+tan¹(x+y/1-xy), xy>1;x, y>0

Asked by joymaibam38 | 26th Dec, 2021, 09:04: AM

Expert Answer:

x > 0, y > 0
rightwards double arrow 0 less than tan to the power of negative 1 end exponent straight x less than straight pi over 2 space space & space space 0 less than tan to the power of negative 1 end exponent straight y less than straight pi over 2
Let space tan to the power of negative 1 end exponent straight x equals straight alpha space and space tan to the power of negative 1 end exponent straight y equals straight beta
rightwards double arrow straight x equals tan space straight alpha space & space tan space straight beta equals straight y
tan open parentheses straight alpha plus straight beta close parentheses equals open parentheses tanα plus tanβ close parentheses open parentheses 1 minus tanα space tanβ close parentheses
equals open parentheses straight x plus straight y close parentheses open parentheses 1 minus xy close parentheses
less than 0
straight pi over 2 less than straight alpha plus straight beta less than straight pi
rightwards double arrow negative straight pi over 2 less than straight alpha plus straight beta minus straight pi less than 0
tan open parentheses straight alpha plus straight beta minus straight pi close parentheses equals negative tan open parentheses straight pi minus open parentheses straight alpha plus straight beta close parentheses close parentheses equals equals tan open parentheses straight alpha plus straight beta close parentheses equals fraction numerator straight x plus straight y over denominator 1 minus xy end fraction
rightwards double arrow 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 equals straight alpha plus straight beta minus straight pi
rightwards double arrow tan to the power of negative 1 end exponent straight x plus tan to the power of negative 1 end exponent straight y equals straight pi plus 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

Answered by Renu Varma | 26th Dec, 2021, 12:25: PM