For 0 2 , the solution(s) of m = 1 6 c o s e c ( + ( m - 1 ) 4 ) c o s e c ( + m 4 ) = 4 2 is (are) [2009]

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) [2009]

Smart Achievers · Accepted Answer

(3, 4)

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

$\Rightarrow \sum_{m = 1}^{6} \frac{\sin \frac{π}{4}}{\sin [θ + \frac{(m - 1) π}{4}] \sin [θ + \frac{m π}{4}]} = 4$

$\Rightarrow \sum_{m = 1}^{6} \frac{\sin [(θ + \frac{m π}{4}) - (θ + \frac{(m - 1) π}{4})]}{\sin (θ + \frac{(m - 1) π}{4}) \sin (θ + \frac{m π}{4})} = 4$

$\Rightarrow \sum_{m = 1}^{6} \frac{[\sin (θ + \frac{m π}{4}) \cos (θ + \frac{(m - 1) π}{4}) - \cos (θ + \frac{m π}{4}) \sin (θ + \frac{(m - 1) π}{4})]}{\sin (θ + \frac{(m - 1) π}{4}) \sin (θ + \frac{m π}{4})} = 4$

$\Rightarrow \sum_{m = 1}^{6} [\cot (θ + \frac{(m - 1) π}{4}) - \cot (θ + \frac{m π}{4})] = 4$

$\Rightarrow [\cot θ - \cot (θ + \frac{π}{4})] + [\cot (θ + \frac{π}{4}) - \cot (θ + \frac{2 π}{4})] + \dots + [\cot (θ + \frac{5 π}{4}) - \cot (θ + \frac{6 π}{4})] = 4$

$\Rightarrow \cot θ - \cot (θ + \frac{3 π}{2}) = 4$ $\Rightarrow \cot θ + \tan θ = 4$

$\Rightarrow \cos^{2} θ + \sin^{2} θ = 4 \sin θ \cos θ$

$\Rightarrow \sin 2 θ = \frac{1}{2}$ $\Rightarrow 2 θ = \frac{π}{6} or \frac{5 π}{6}$ $\Rightarrow θ = \frac{π}{12} or \frac{5 π}{12}$

Solution

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) [2009]

1 $\frac{π}{4}$

2 $\frac{π}{6}$

3 $\frac{π}{12}$

4 $\frac{5 π}{12}$

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Solution

Q. For 0<θ<π2, the solution(s) of ∑m=16cosec(θ+(m-1)π4)cosec(θ+mπ4)=42 is (are) [2009]

1 π4

2 π6

3 π12