<s>[INST] $ \sum_{m=1}^{6}\frac{1}{\sin \left\{\theta+\left(m-1\right)\cdot \frac{\pi}{4}\right\} \sin \left\{\theta+m\cdot \frac{\pi}{4}\right\}} = 4\sqrt{2}$ If $\displaystyle 0 < \theta < \frac{\pi}{2}$ and $\displaystyle \sum_{m=1}^{6}\frac{1}{\sin \left\{\theta+\left(m-1\right)\cdot \frac{\pi}{4}\right\}\cdot \sin \left\{\theta+m\cdot \frac{\pi}{4}\right\}} = 4\sqrt{2}$. Then value of $\theta=.$
$\bf{My\; Try::}$ Here the question is given in the form of $\displaystyle \frac{1}{\sin A \cdot \sin B}$. Then we can write it as 
$$\displaystyle \Rightarrow \frac{1}{\sin A \cdot \sin B} = \frac{1}{\sin (B-A)}\cdot \left[\frac{\sin (B-A)}{\sin A\cdot \sin B}\right] $$
$$\displaystyle \Rightarrow \frac{1}{\sin(B-A)}\cdot \left[\frac{\cos A\cdot \sin B-\sin A\cdot \cos B}{\sin A \cdot \sin B}\right]$$
$$\displaystyle = \frac{1}{\sin (B-A)}\cdot \left[\cot A-\cot B\right]$$
Now Let $\displaystyle \left\{\theta+(m-1)\cdot \frac{\pi}{4}\right\}=A$ and $\displaystyle \left\{\theta+m\cdot \frac{\pi}{4}\right\}=B.$
So We Get $\displaystyle \sum_{m=1}^{6}\frac{1}{\sin A \cdot \sin B}=\sum_{m=1}^{6}\sqrt{2}\left[\cot A-\cot B\right] = 4\sqrt{2}.$
So we get $$\displaystyle \sum_{m=1}^{6}\left[\cot \left\{\theta+(m-1)\cdot \frac{\pi}{4}\right\}-\cot \left\{\theta+m\cdot \frac{\pi}{4}\right\}\right]=4.$$
Now after expanding we get $$\displaystyle \cot \theta - \cot \left(\theta+\frac{6\pi}{4}\right) = 4.\Rightarrow \cot \theta - \cot \left(\theta+\frac{2\pi}{3}\right)=4$$
Again after simplifying we get $$\displaystyle \cot \theta - \left(\frac{\cot \theta-\sqrt{3}}{1-\sqrt{3}\cot \theta}\right)=4\Rightarrow \frac{\sqrt{3}-\sqrt{3}\cot^2 \theta}{1-\sqrt{3}\cot \theta} = 4.$$
So We get $$\displaystyle \sqrt{3}-\sqrt{3}\cot^2 \theta = 4-4\sqrt{3}\cot \theta.\Rightarrow \sqrt{3}\cot^2 \theta - 4\sqrt{3}\cot \theta+(4-\sqrt{3})=0$$.
Now Let $\cot \theta = y\;,$ Then equation is $\sqrt{3}y^2-4\sqrt{3}y+(4-\sqrt{3}) = 0.$
We Get $$\displaystyle y = \cot \theta = \frac{2\pm \sqrt{15-4\sqrt{3}}}{3}$$
But I did not understand how can i get value of $\theta.$
Help me, Thanks
 [/INST] HINT:
Can you try tis way?
$$\cot\theta-\cot\left(\frac{2\pi}3+\theta\right)=\frac{\sin\dfrac{2\pi}3}{\cos\theta\cdot\cos\left(\frac{2\pi}3+\theta\right)}=\frac{\sqrt3}{2\cos\theta\cdot\cos\left(\frac{2\pi}3+\theta\right)}$$
$$=\frac{\sqrt3}{\cos\dfrac{2\pi}3+\cos\left(\dfrac{2\pi}3+2\theta\right)}$$
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