Inverse Trigonometric Functions

Find the value of the following: - Wired Faculty

Question

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

Solution

We know that:
$sin to the power of negative 1 end exponent fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction equals 2 tan to the power of negative 1 end exponent straight x space for space open vertical bar straight x close vertical bar less or equal than 1 space... left parenthesis 1 right parenthesis cos to the power of negative 1 end exponent fraction numerator 1 minus straight y squared over denominator 1 plus straight y squared end fraction equals 2 tan to the power of negative 1 end exponent straight y space for space straight y greater than 0 space... left parenthesis 2 right parenthesis therefore space space sin to the power of negative 1 end exponent fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction plus cos to the power of negative 1 end exponent fraction numerator 1 minus straight y squared over denominator 1 plus straight y squared end fraction equals 2 tan to the power of negative 1 end exponent straight x plus 2 tan to the power of negative 1 end exponent straight y. rightwards double arrow space tan 1 half open square brackets sin to the power of negative 1 end exponent fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction plus cos to the power of negative 1 end exponent fraction numerator 1 minus straight y squared over denominator 1 plus straight y squared end fraction close square brackets$
$equals tan 1 half left parenthesis 2 tan to the power of negative 1 end exponent straight x plus 2 tan to the power of negative 1 end exponent straight y right parenthesis equals tan left parenthesis tan to the power of negative 1 end exponent straight x plus tan to the power of negative 1 end exponent straight y right parenthesis equals tan open parentheses tan to the power of negative 1 end exponent fraction numerator straight x plus straight y over denominator 1 minus xy end fraction close parentheses space space open square brackets because tan to the power of negative 1 end exponent straight x plus tan to the power of negative 1 end exponent straight y equals tan to the power of negative 1 end exponent fraction numerator straight x plus straight y over denominator 1 minus xy end fraction comma space for space xy less than 1 close square brackets equals fraction numerator straight x plus straight y over denominator 1 minus xy end fraction$

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