Question

Find the value of tan $1 half open square brackets sin to the power of negative 1 end exponent fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction plus cos to the power of negative 1 end exponent fraction numerator 1 minus straight y squared over denominator 1 plus straight y squared end fraction close square brackets comma space open vertical bar straight x close vertical bar less than 1 comma space straight y greater than 0 space and space straight x space straight y space less than 1$

Solution

straight L. straight H. straight S. space equals 2 space tan to the power of negative 1 end exponent 1 half open square brackets sin to the power of negative 1 end exponent fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction plus cos to the power of negative 1 end exponent fraction numerator 1 minus straight y squared over denominator 1 plus straight y squared end fraction close square brackets equals tan open square brackets 1 half sin to the power of negative 1 end exponent fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction plus 1 half cos fraction numerator 1 minus straight y squared over denominator 1 plus straight y squared end fraction close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space tan space left square bracket space tan to the power of negative 1 end exponent space straight x plus tan to the power of negative 1 end exponent space straight y space right square bracket space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space tan space open square brackets 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 close square brackets equals fraction numerator straight x plus straight y over denominator 1 minus xy end fraction space where space straight x space straight y space less than 1 space space space space space space space space space space space

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