Solving trigonometric equation involving summation For $ 0 <\theta<\frac{\pi}{2}$ find the solution of
$$\sum\limits_{m=1}^{6}\csc\left(\theta+\frac{(m-1)\pi}{4}\right)\cdot\csc\left(\theta+\frac{m\pi}{4}\right)=4\sqrt{2}$$
I thought of solving this as the angles form an A.P , But the given sum does not
come under any standard type such as the sum of the sines or cosines of
the angles in an A.P.So I am unable to proceed further.

The left side of the equation can be rewritten as:
$$ \Delta = \sum_{1 3 5} \csc\left(\theta+\frac{m\pi}{4}\right) \cdot \left( \csc\left(\theta+\frac{(m-1)\pi}{4}\right) +  \csc\left(\theta+\frac{(m+1)\pi}{4}\right) \right) $$
Now using the formulas
$$ \sin u +\sin v = 2\sin\left(\frac {u + v} 2\right) \cdot \cos \left(\frac {u - v} 2\right)$$
and
$$ \sin u \sin v = \frac 1 2 \left(\cos (u - v) - \cos (u + v)\right)$$
we have
$$ \begin{aligned}\csc\left(\theta+\frac{(m-1)\pi}{4}\right) +  \csc\left(\theta+\frac{(m+1)\pi}{4}\right) &= \frac {\sin\left(\theta+\frac{(m-1)\pi}{4}\right) +  \sin\left(\theta+\frac{(m+1)\pi}{4}\right)} {\sin\left(\theta+\frac{(m-1)\pi}{4}\right) \cdot \sin\left(\theta+\frac{(m+1)\pi}{4}\right)}\\ &= \frac {4 \sin\left(\theta + \frac {m\pi} {4}\right) cos\left( \frac \pi 4\right)}{\cos\left(\frac \pi 2 \right) - \cos\left(2\theta + \frac {m\pi} {2} \right)}\end{aligned}$$
So $\Delta$ becomes
$$\Delta = -2\sqrt 2\sum_{1 3 5} \frac 1 {\cos\left(2\theta + \frac {m\pi} {2} \right)} = -2\sqrt 2 \left( -\frac 1 {\sin 2\theta} +\frac 1 {\sin 2\theta} -\frac 1 {\sin 2\theta}\right) = 2\sqrt 2 \frac 1 {\sin 2\theta}$$
I hope there are no typos... ;)
