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Determinant

For 0 < θ <...

Question

For $0 < θ < \frac{π}{2}$ , the solution of $6 \sum m = 1 cos e c (θ + \frac{(m - 1) π}{4}) cos e c (θ + \frac{m π}{4}) = 4 \sqrt{2}$ is/are

A

\frac{π}{4}

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B

\frac{π}{6}

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C

\frac{π}{12}

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D

\frac{5 π}{12}

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Solution

The correct options are
C $\frac{π}{12}$
D $\frac{5 π}{12}$
We have,

$6 \sum m = 1 cos e c [θ + \frac{(m - 1) π}{4}] cos e c (θ + \frac{m π}{4}) = 4 \sqrt{2}$

$6 \sum m = 1 \frac{1}{sin [θ + \frac{(m - 1) π}{4}] sin (θ + \frac{m π}{4})} = 4 \sqrt{2}$

$6 \sum m = 1 \frac{1}{sin [θ + \frac{(m - 1) π}{4}] sin (θ + \frac{m π}{4})} = 4 \sqrt{2}$

$\frac{1}{sin θ sin (θ + \frac{π}{4})} + \frac{1}{sin (θ + \frac{π}{4}) sin (θ + \frac{π}{2})} + . . . . . + \frac{1}{sin (θ + \frac{5 π}{4}) sin (θ + \frac{3 π}{2})} = 4 \sqrt{2}$

$\frac{1}{sin \frac{π}{4}} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{sin (θ + \frac{π}{4} - θ)}{sin θ sin (θ + \frac{π}{4})} + \frac{sin (θ + \frac{π}{2} - (θ + \frac{π}{4}))}{sin (θ + \frac{π}{4}) sin (θ + \frac{π}{2})} + . . . . . + \frac{sin (θ + \frac{3 π}{2} - (θ + \frac{5 π}{4}))}{sin (θ + \frac{5 π}{4}) sin (θ + \frac{3 π}{2})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ = 4 \sqrt{2}$

$\sqrt{2} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{sin (θ + \frac{π}{4}) cos θ - cos (θ + \frac{π}{4}) sin θ}{sin θ sin (θ + \frac{π}{4})} + \frac{sin (θ + \frac{π}{2}) cos (θ + \frac{π}{4}) - cos (θ + \frac{π}{2}) sin (θ + \frac{π}{4})}{sin (θ + \frac{π}{4}) sin (θ + \frac{π}{2})} + . . . . . ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ = 4 \sqrt{2}$

$[cot θ - cot (θ + \frac{π}{4}) + cot (θ + \frac{π}{4}) - cot (θ + \frac{π}{2}) + . . . . . + cot (θ + \frac{5 π}{4}) - cot (θ + \frac{3 π}{2})] = 4$

$cot θ - cot (θ + \frac{π}{2}) = 4$

$cot θ + tan θ = 4$

${tan}^{2} θ - 4 tan θ + 1 = 0$

$tan θ = \frac{4 \pm \sqrt{16 - 4 \times 1 \times 1}}{2}$

$tan θ = \frac{4 \pm \sqrt{16 - 4}}{2}$

$tan θ = \frac{4 \pm \sqrt{12}}{2}$

$tan θ = \frac{4 \pm 2 \sqrt{3}}{2}$

$tan θ = 2 \pm \sqrt{3}$

$θ = \frac{π}{12}$ or $\frac{5 π}{12}$

Hence, the value of $θ$ is $\frac{π}{12}$ or $\frac{5 π}{12}$

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Similar questions

0 < θ < \frac{π}{2},

the solution (s) of

\sum_{m = 1}^{6} c o s e c (θ + \frac{(m - 1) π}{4}) c o s e c (θ + \frac{m π}{4}) = 4 \sqrt{2}

0 < θ < \frac{π}{2}

, the solution(s) of

\sum_{m = 1}^{6} c o s e c (θ + \frac{(m - 1) π}{4}) c o s e c (θ + \frac{m π}{4}) = 4 \sqrt{2}

\int_{0}^{3} \frac{3 x + 1}{x^{2} + 9} d x =

\frac{π}{12} + \log (2 \sqrt{2})

\frac{π}{2} + \log (2 \sqrt{2})

\frac{π}{6} + \log (2 \sqrt{2})

\frac{π}{3} + \log (2 \sqrt{2})

Q. Let f : (-1, 1)

\to

R be such that f

(cos 4 θ)

\frac{2}{2 - {sec}^{2} θ}

θ \in (0, \frac{π}{4}) \cup (\frac{π}{4}, \frac{π}{2}) .

Then, the value(s) of

f (\frac{1}{3})

0 < θ < \frac{π}{2},

(s)

6 \sum m - 1

(0 + \frac{(m - 1) π}{4})

(0 + \frac{m π}{4}) = 4 \sqrt{2}

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