∑ m = 1 6 cosec ⁡ θ + ( m − 1 ) π 4 ∣ cosec ⁡ θ + m π 4 ∣ = 4 2 ⇒ ∑ m = 1 6 sin ⁡ π 4 sin ⁡ θ + ( m − 1 ) π ] 4 sin ⁡ θ + m + π 4 = 4 ⇒ ∑ m = 1 6 sin ⁡ θ + m π 4 − θ + ( m − 1 ) π 4 sin ⁡ θ + ( m − 1 ) π 4 sin ⁡ θ + m π 4 = 4 ⇒ ∑ m = 1 6 sin ⁡ θ + m π 4 cos ⁡ θ + ( m − 1 ) π 4 - c o s θ + m π 4 sin ⁡ θ + ( m − 1 ) π 4 sin ⁡ θ + ( m − 1 ) π 4 sin ⁡ θ + m π 4 = 4 ⇒ ∑ m = 1 6 cot ⁡ θ + ( m − 1 ) π 4 − cot ⁡ θ + m π 4 = 4 ⇒ cot ⁡ θ − cot ⁡ θ + π 4 + cot ⁡ θ + π 4 − cot ⁡ θ + 2 π 4 + … + cot ⁡ θ + 5 π 4 − cot ⁡ θ + 6 π 4 = 4 ⇒ cot ⁡ θ − cot ⁡ θ + 3 π 2 = 4 ⇒ cot ⁡ θ + tan ⁡ θ = 4 ⇒ cos 2 ⁡ θ + sin 2 ⁡ θ = 4 sin ⁡ θ cos ⁡ θ ⇒ sin ⁡ 2 θ = 1 2 ⇒ 2 θ = π 6 or 5 π 6 ⇒ θ = π 12 or 5 π 12

For 0<θ<π2, the solution (s) of ∑m=16 cosec⁡θ+(m−1)π4cosec⁡θ+mπ4=42 is (are)

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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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$\frac{π}{4}$

$\frac{π}{12}$

$\frac{π}{6}$

$\frac{5 π}{12}$

answer is C, D.

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Detailed Solution

$\begin{array}{l} \sum_{m = 1}^{6} cosec [θ + \frac{(m - 1) π}{4} ∣] cosec [θ + \frac{m π}{4} ∣] = 4 \sqrt{2} \\ \Rightarrow \sum_{m = 1}^{6} \frac{\sin \frac{π}{4}}{\sin [θ + \frac{(m - 1) π]}{4} \sin [θ + \frac{m + π}{4}]} = 4 \\ \Rightarrow \sum_{m = 1}^{6} \frac{\sin [(θ + \frac{m π}{4}) - (θ + \frac{(m - 1) π}{4})]}{\sin (θ + \frac{(m - 1) π}{4}) \sin (θ + \frac{m π}{4})} = 4 \\ \Rightarrow \sum_{m = 1}^{6} [\frac{\sin (θ + \frac{m π}{4}) \cos (θ + \frac{(m - 1) π}{4}) - c o s (θ + \frac{m π}{4}) \sin (θ + \frac{(m - 1) π}{4})}{\sin (θ + \frac{(m - 1) π}{4}) \sin (θ + \frac{m π}{4})} = 4 \\ \Rightarrow \sum_{m = 1}^{6} [\cot (θ + \frac{(m - 1) π}{4}) - \cot (θ + \frac{m π}{4})] = 4 \\ \Rightarrow [\cot θ - \cot (θ + \frac{π}{4})] + [\cot (θ + \frac{π}{4}) - \cot (θ + \frac{2 π}{4})] \\ + \dots + [\cot (θ + \frac{5 π}{4}) - \cot (θ + \frac{6 π}{4})] = 4 \end{array}$

$\begin{array}{l} \Rightarrow \cot θ - \cot (θ + \frac{3 π}{2}) = 4 \Rightarrow \cot θ + \tan θ = 4 \\ \Rightarrow \cos^{2} θ + \sin^{2} θ = 4 \sin θ \cos θ \\ \Rightarrow \sin 2 θ = \frac{1}{2} \Rightarrow 2 θ = \frac{π}{6} or \frac{5 π}{6} \\ \Rightarrow θ = \frac{π}{12} or \frac{5 π}{12} \end{array}$