Question

Solve, the following equations;

$tan to the power of negative 1 end exponent open parentheses fraction numerator 1 minus x over denominator 1 plus x end fraction close parentheses equals 1 half tan to the power of negative 1 end exponent space x comma space left parenthesis x space greater than space 0 right parenthesis$

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Solution

The given equation is

tan to the power of negative 1 end exponent open parentheses fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction close parentheses equals 1 half space tan to the power of negative 1 end exponent space straight x space space space space space space space space space space or space space space 2 space tan to the power of negative 1 end exponent open parentheses fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction close parentheses equals tan to the power of negative 1 end exponent straight x or space space space space tan to the power of negative 1 end exponent space open square brackets fraction numerator 2 open parentheses begin display style fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction end style close parentheses over denominator 1 minus open parentheses begin display style fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction end style close parentheses squared end fraction close square brackets space tan to the power of negative 1 end exponent straight x space space space space space space space space space space space space space space space space space space space space open square brackets because space 2 space tan space straight theta equals space tan to the power of negative 1 end exponent space fraction numerator 2 straight theta over denominator 1 minus straight theta squared end fraction close square brackets or space space space space tan to the power of negative 1 end exponent open square brackets fraction numerator 2 left parenthesis 1 minus straight x right parenthesis left parenthesis 1 plus straight x right parenthesis over denominator left parenthesis 1 plus straight x right parenthesis squared minus left parenthesis 1 minus straight x right parenthesis squared end fraction close square brackets equals space tan to the power of negative 1 end exponent space straight x therefore space space space tan to the power of negative 1 end exponent space open square brackets fraction numerator 2 left parenthesis 1 minus straight x squared right parenthesis over denominator 4 space straight x end fraction close square brackets equals tan to the power of negative 1 end exponent space straight x therefore space space space space fraction numerator 2 left parenthesis 1 minus straight x squared right parenthesis over denominator 4 straight x end fraction equals straight x space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space space space 1 minus straight x squared equals 2 space straight x squared therefore space space space space space 3 space straight x squared equals 1 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 rightwards double arrow space space space space space straight x squared equals 1 third therefore space space space space space space space space space space space straight x equals fraction numerator 1 over denominator square root of 3 end fraction

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