Inverse Trigonometric Functions

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Question

Solve : $tan to the power of negative 1 end exponent open parentheses fraction numerator 1 plus x over denominator 1 minus x end fraction close parentheses equals straight pi over 4 plus tan to the power of negative 1 end exponent comma space x comma space o less than x less than 1.$

Solution

The given equation is

tan to the power of negative 1 end exponent open parentheses fraction numerator 1 plus x over denominator 1 minus x end fraction close parentheses equals straight pi over 4 plus tan to the power of negative 1 end exponent x space space space space space space space space space space space space space space o t space tan to the power of negative 1 end exponent open parentheses fraction numerator 1 plus x over denominator 1 minus x end fraction close parentheses minus tan to the power of negative 1 end exponent space x space equals space straight pi over 4 tan to the power of negative 1 end exponent open square brackets fraction numerator begin display style fraction numerator 1 plus x over denominator 1 minus x end fraction end style minus x over denominator 1 plus open parentheses begin display style fraction numerator 1 plus x over denominator 1 minus x end fraction end style close parentheses left parenthesis x right parenthesis end fraction close square brackets equals straight pi over 4 space space space space space space space space space space space space space space space space space space open square brackets because space space tan to the power of negative 1 end exponent x minus tan to the power of negative 1 end exponent space y space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator x minus y over denominator 1 plus x space y end fraction close parentheses close square brackets therefore space space space space space tan to the power of negative 1 end exponent open square brackets fraction numerator 1 plus x minus x left parenthesis 1 minus x right parenthesis over denominator 1 minus x plus x left parenthesis 1 plus x right parenthesis end fraction close square brackets equals straight pi over 4 therefore space space space space space space space tan to the power of negative 1 end exponent left parenthesis 1 right parenthesis equals straight pi over 4 therefore space space space space space space space straight pi over 4 equals straight pi over 4

∴ equation has infinitely many solutions.

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Inverse Trigonometric Functions

Solve : $tan to the power of negative 1 end exponent open parentheses fraction numerator 1 plus x over denominator 1 minus x end fraction close parentheses equals straight pi over 4 plus tan to the power of negative 1 end exponent comma space x comma space o less than x less than 1.$

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