For 0-2- the solution s of -m-16cosec - - m-1-4 cosec - - m-4 - 4-2 is-are

For 0 lt; θ lt; π/2 ∑_m=1^6cosec ( θ + (m 1)π4 ) cosec ( θ + mπ4 ) = 4√(2) ⇒∑_m=1^61sin ( θ + (m 1)π4 )sin ( θ + mπ4 ) = 4√(2) ⇒∑_m=1^6sinπ4sin ( θ + (m ...

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For 0<θ<π2, t...

Question

For $0 < θ < \frac{π}{2},$ the solution (s) of
$\sum_{m = 1}^{6} cosec (θ + \frac{(m - 1) π}{4}) cosec (θ + \frac{m π}{4}) = 4 \sqrt{2}$ is/are

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Solution

For $0 < θ < \frac{π}{2}$
$\sum_{m = 1}^{6} cosec (θ + \frac{(m - 1) π}{4}) cosec (θ + \frac{m π}{4}) = 4 \sqrt{2}$
$\Rightarrow \sum_{m = 1}^{6} \frac{1}{sin (θ + \frac{(m - 1) π}{4}) sin (θ + \frac{m π}{4})} = 4 \sqrt{2}$
$\Rightarrow \sum_{m = 1}^{6} \frac{sin \frac{π}{4}}{sin (θ + \frac{(m - 1) π}{4}) sin (θ + \frac{m π}{4})} = 4$
$\Rightarrow \sum_{m = 1}^{6} \frac{sin [θ + \frac{m π}{4} - (θ + \frac{(m - 1) π}{4})]}{sin (θ + \frac{(m - 1) π}{4}) sin (θ + \frac{m π}{4})} = 4$
$\Rightarrow \sum_{m = 1}^{6} \frac{sin (θ + \frac{m π}{4}) cos (θ + \frac{(m - 1) π}{4}) - cos (θ + \frac{m π}{4}) sin (θ + \frac{(m - 1) π}{4})}{sin (θ + \frac{(m - 1) π}{4}) sin (θ + \frac{m π}{4})} = 4$
$\Rightarrow \sum_{m = 1}^{6} \frac{cos (θ + \frac{(m - 1) π}{4})}{sin (θ + \frac{(m - 1) π}{4})} - \frac{cos (θ + \frac{m π}{4})}{sin (θ + \frac{m π}{4})} = 4$
$\Rightarrow \sum_{m = 1}^{6} [cot (θ + \frac{(m - 1) π}{4}) - cot (θ + \frac{m π}{4})] = 4$
Put the value of $m$
$\Rightarrow cot (θ) - cot (θ + \frac{π}{4}) + cot (θ + \frac{π}{4})$
$- cot (θ + \frac{2 π}{4}) + . . . + cot (θ + \frac{5 π}{4}) - cot (θ + \frac{6 π}{4}) = 4$
$\Rightarrow cot θ - cot (\frac{3 π}{2} + θ) = 4$
$\Rightarrow cot θ + tan θ = 4$
$\Rightarrow {tan}^{2} θ - 4 tan θ + 1 = 0$
$\Rightarrow (tan θ - 2)^{2} - 3 = 0$
$\Rightarrow (tan θ - 2 + \sqrt{3}) (tan θ - 2 - \sqrt{3}) = 0$
$\Rightarrow tan θ = 2 - \sqrt{3} or tan θ = 2 + \sqrt{3}$
$\Rightarrow θ = \frac{π}{12}; θ = \frac{5 π}{12}$
$∵ θ \in (0, \frac{θ}{2})$
Hence the correct answer are Option c and Option d.

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Similar questions

Q. Assertion :The value of

θ, 0 < θ < \frac{π}{2},

satisfying the equation

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

are

\frac{π}{12}

and

\frac{5 π}{12}

Reason: If

tan θ + cot θ = 4, 0 < θ < \frac{π}{2},

then

θ = \frac{π}{12}

\frac{5 π}{12}

Q. For

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}

is/are

Q. For

0 < θ < \frac{π}{2},

the solution

(s)

6 \sum m - 1

cosec

(0 + \frac{(m - 1) π}{4})

cosec

(0 + \frac{m π}{4}) = 4 \sqrt{2}

is (are)

Q. For

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}

is(are)

Q. For

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}

is/are

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For 0<θ<π2, the solution (s) of ∑6m=1cosec (θ+(m−1)π4)cosec (θ+mπ4)=4√2 is/are

For $0 < θ < \frac{π}{2},$ the solution (s) of
$\sum_{m = 1}^{6} cosec (θ + \frac{(m - 1) π}{4}) cosec (θ + \frac{m π}{4}) = 4 \sqrt{2}$ is/are