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Inverse Trigonometric Functions

Question
CBSEENMA12032738
Wired Faculty App

Prove that tan to the power of negative 1 end exponent space straight t plus tan to the power of negative 1 end exponent fraction numerator 2 space straight l over denominator 1 minus straight t squared end fraction equals tan to the power of negative 1 end exponent fraction numerator 3 straight t minus straight t cubed over denominator 1 minus 3 straight t squared end fraction. straight t greater than 0

Solution
straight L. straight H. straight S. equals tan to the power of negative 1 end exponent space straight t plus tan to the power of negative 1 end exponent fraction numerator 2 straight t over denominator 1 minus straight t squared end fraction equals tan to the power of negative 1 end exponent open square brackets fraction numerator straight t plus begin display style fraction numerator 2 straight t over denominator 1 minus straight t squared end fraction end style over denominator 1 minus straight t cross times begin display style fraction numerator 2 space straight t over denominator 1 minus straight t squared end fraction end style end fraction close square brackets space space space space space space space space space space equals tan to the power of negative 1 end exponent open square brackets fraction numerator straight t left parenthesis 1 minus straight t squared right parenthesis plus 2 straight t over denominator 1 minus straight t squared minus 2 straight t squared end fraction close square brackets space space space space space space space space space space equals tan to the power of 1 open square brackets fraction numerator straight t minus straight t cubed plus 2 straight t over denominator 1 minus 3 straight t squared end fraction close square brackets equals tan to the power of negative 1 end exponent open parentheses fraction numerator 3 straight t minus straight t cubed over denominator straight t minus 3 straight t squared end fraction close parentheses equals straight R. straight H. straight S.

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