For 0 lt- lt-2- the solution -s- of- -m-1-6 cosec - - - -m-1-4 - cosec - - - m-4 - - 4-2- is-are

The correct options are C D For 0 lt; θ lt; π/2 ∑_m=1^6 cosec ( θ + (m 1)π/4 ) cosec ( θ + mπ/4 ) = 4√(2) ⇒∑_m=1^61/sin ( θ + (m 1)π/4 )sin ( θ + mπ/4 ) = ...

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Byju's Answer

Standard XI

Mathematics

Principal Solution of Trigonometric Equation

For 0 lt;θ ...

Question

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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Solution

The correct options are
C
D
For $0 < θ < \frac{π}{2}$
$\sum_{m = 1}^{6} c o s e c (θ + \frac{(m - 1) π}{4}) c o s e c (θ + \frac{m π}{4}) = 4 \sqrt{2}$
$\Rightarrow \sum_{m = 1}^{6} \frac{1}{sin (θ + \frac{(m - 1) π}{4}) sin (θ + \frac{m π}{4})} = 4 \sqrt{2}$
$\Rightarrow \sum_{m = 1}^{6} \frac{sin [θ + \frac{m π}{4} - (θ + \frac{(m - 1) π}{4})]}{sin (θ + \frac{(m - 1) π}{4}) sin (θ + \frac{m π}{4})} = 4$
$\Rightarrow \sum_{m = 1}^{6} [cot (θ + \frac{(m - 1) π}{4}) - cot (θ + \frac{m π}{4})] = 4$
$\Rightarrow cot (θ) - cot (θ + \frac{π}{4}) + cot (θ + \frac{π}{4})$
$- cot (θ + \frac{2 π}{4}) + . . . + cot (θ + \frac{5 π}{4}) - cot (θ + \frac{6 π}{4}) = 4$
$\Rightarrow cot θ - cot (\frac{3 π}{2} + θ) = 4$
$\Rightarrow cot θ + tan θ = 4$
$\Rightarrow {tan}^{2} θ - 4 tan θ + 1 = 0$
$\Rightarrow (tan θ - 2)^{2} - 3 = 0$
$\Rightarrow (tan θ - 2 + \sqrt{3}) (tan θ - 2 - \sqrt{3}) = 0$
$\Rightarrow tan θ = 2 - \sqrt{3} o r tan θ = 2 + \sqrt{3}$
$\Rightarrow θ = \frac{π}{12}; θ = \frac{5 π}{12}$
$∵ θ \in (0, \frac{θ}{2})$

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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

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