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Question

If fraction numerator sin cubed space theta minus cos cubed space theta over denominator sin space theta minus cos space theta end fraction minus fraction numerator cos begin display style space end style theta over denominator square root of 1 plus cot 2 space theta end root end fraction minus 2 tan space theta cot space theta equals negative 1,then

  1. theta element of open parentheses 0 comma pi over 2 close parentheses
  2. theta element of open parentheses 0 comma pi over 2 close parentheses
  3. theta element of open parentheses pi comma fraction numerator 3 pi over denominator 2 end fraction close parentheses
  4. theta element of open parentheses fraction numerator 3 pi over denominator 2 end fraction comma 2 pi close parentheses

hintHint:

we have given trigonometry function. Which is
fraction numerator sin cubed space theta minus cos cubed space theta over denominator sin space theta minus cos space theta end fraction minus fraction numerator cos begin display style space end style theta over denominator square root of 1 plus cot 2 space theta end root end fraction minus 2 tan space theta cot space theta equals negative 1
We have to find where θ is belongs. Solve the function and find the answer.

The correct answer is: theta element of open parentheses fraction numerator 3 pi over denominator 2 end fraction comma 2 pi close parentheses


    Here, We have to find where θ is belongs.
    Firstly, we have given
    fraction numerator sin cubed space theta minus cos cubed space theta over denominator sin space theta minus cos space theta end fraction minus fraction numerator cos begin display style space end style theta over denominator square root of 1 plus cot 2 space theta end root end fraction minus 2 tan space theta cot space theta equals negative 1   ---(1)
    fraction numerator sin cubed space theta minus cos cubed space theta over denominator sin space theta minus cos space theta end fraction minus fraction numerator cos begin display style space end style theta over denominator square root of 1 plus cot 2 space theta end root end fraction minus 2 equals negative 1            [since, fraction numerator tan space theta space equals space 1 over denominator c o t space theta end fraction]
    We know that,
    a cubed minus b cubed equals left parenthesis a minus b right parenthesis left parenthesis a squared plus a b plus b squared right parenthesis
    1 plus fraction numerator sin theta cos theta minus cos theta over denominator open parentheses square root of left parenthesis 1 plus 2 cos squared theta – 1 end root close parentheses end fraction – 2 equals negative 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 space space left square bracket c o t 2 theta equals 2 c o s squared theta – 1 right square bracket
1 plus fraction numerator sin theta cos theta minus cos theta over denominator open parentheses square root of 2 left parenthesis cos squared theta end root close parentheses end fraction – 2 equals negative 1
1 plus s i n theta c o s theta minus fraction numerator cos theta over denominator square root of 2 c o s theta – 2 end root end fraction equals negative 1
1 plus s i n theta c o s theta – fraction numerator 1 over denominator square root of 2 – 2 end root end fraction equals negative 1
s i n theta c o s theta – fraction numerator 1 over denominator square root of 2 end fraction equals 0
s i n theta c o s theta equals fraction numerator 1 over denominator square root of 2 end fraction
    For every θ for sin and cos is straight pi over 4 . so for But θ not possible in second quadrant because cos is negative,
    Then θ always belongs to ( 0 , straight pi over 2 ).
    Therefore the correct answer is θ ∈ ( 0 , straight pi over 2).
    fraction numerator sin cubed space theta minus cos cubed space theta over denominator sin space theta minus cos space theta end fraction minus fraction numerator cos begin display style space end style theta over denominator square root of 1 plus cot 2 space theta end root end fraction minus 2 tan space theta cot space theta equals negative 1

    In this question, we have given equation, where we have to find where the θ belongs. For all value of sin θ cos θ = fraction numerator 1 over denominator square root of 2 end fraction so θ always lies between ( 0 , straight pi over 2 ) .

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