Maths-
General
Easy

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 open parentheses 1 plus cot squared space theta close parentheses end root end fraction minus 2 tan space theta cot space theta equals negative 1 comma theta element of left square bracket 0 comma 2 pi right square bracket comma,then

  1. theta element of open parentheses 0 comma pi over 2 close parentheses minus open curly brackets pi over 4 close curly brackets
  2. theta element of open parentheses pi over 2 comma pi close parentheses minus open curly brackets fraction numerator 3 pi over denominator 4 end fraction close curly brackets
  3. theta element of open parentheses pi comma fraction numerator 3 pi over denominator 2 end fraction close parentheses minus open curly brackets fraction numerator 5 pi over denominator 4 end fraction close curly brackets
  4. theta element of left parenthesis 0 comma pi right parenthesis minus open curly brackets pi over 4 comma pi over 2 close curly brackets

hintHint:

In this question, we have given trigonometry function. Which isfraction numerator left parenthesis sin cubed theta minus cos cubed theta right parenthesis over denominator left parenthesis s i n theta minus c o s theta right parenthesis end fraction minus fraction numerator cos theta over denominator square root of left parenthesis 1 plus c o t squared theta right parenthesis end root end fraction minus 2 t a n theta c o t theta equals negative 1and belongs to [ 0, 2 π]. We have to find where θ is belongs. Solve the function and find the answer.

The correct answer is: theta element of open parentheses pi over 2 comma pi close parentheses minus open curly brackets fraction numerator 3 pi over denominator 4 end fraction close curly brackets


    Here , we have to find where θ lies in this equation.
    Firstly, we have given
    fraction numerator left parenthesis sin cubed theta minus cos cubed theta right parenthesis over denominator left parenthesis s i n theta minus c o s theta right parenthesis end fraction minus fraction numerator cos theta over denominator square root of left parenthesis 1 plus c o t squared theta right parenthesis end root end fraction minus 2 t a n theta c o t theta equals negative 1.......... left parenthesis 1 right parenthesis
fraction numerator left parenthesis sin cubed theta minus cos cubed theta right parenthesis over denominator left parenthesis s i n theta minus c o s theta right parenthesis end fraction minus fraction numerator cos theta over denominator left parenthesis 1 plus c o t squared theta right parenthesis end fraction minus 2 equals negative 1 space left square bracket s i n c e comma t a n theta equals fraction numerator 1 over denominator c o t theta end fraction right square bracket
    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
    fraction numerator left parenthesis sin cubed theta minus cos cubed theta right parenthesis over denominator left parenthesis s i n theta minus c o s theta right parenthesis end fraction equals left parenthesis s i n theta minus c o s theta right parenthesis fraction numerator left parenthesis sin squared theta plus sin theta cos theta plus cos squared theta right parenthesis over denominator left parenthesis s i n theta minus c o s theta right parenthesis end fraction
    = 1 + 2 sin θ cos θ
    So, we can write, in eq (1)
    fraction numerator 1 space plus space space sin space theta space cos space theta space minus space cos space theta over denominator fraction numerator left parenthesis sin squared theta plus cos to the power of theta right parenthesis over denominator sin squared theta end fraction 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 open square brackets c o t squared theta equals fraction numerator cos squared theta over denominator sin squared theta end fraction close square brackets
    sin θ cos θ - |sin θ| cos θ = 0
    cos θ ( sin θ - | sin θ| ) = 0
    hence, sin θ = | sin θ |
    which is always true for sinx ≥ 0 otherwise it is true for x=0,straight pi over 2,fraction numerator 3 straight pi over denominator 2 end fraction,2π
    Therefore, since we also need x≠straight pi over 2,fraction numerator 3 straight pi over denominator 2 end fraction for tanx and x≠0,2π for cotx all the solutions are given by x∈(0,π) with x≠straight pi over 4and x≠straight pi over 2.
    The correct answer is θ ϵ (0 ,π)- { straight pi over 4 ,straight pi over 2}

    In this question, we have to find the where is θ lies. Here, always true for sinx ≥ 0 otherwise it is true for x=0,straight pi over 2,fraction numerator 3 straight pi over denominator 2 end fraction,2π and since we also need x≠straight pi over 2,fraction numerator 3 straight pi over denominator 2 end fraction for tanx and x≠0,2π for cotx all the solutions.

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