Mathematics - Trigonometric Equation Question with Solution | TestHub

For $0<\theta <\frac{\pi }{2}$, $$\begin{array}{r}{\sum }_{m=1}^6\mathrm{cosec}\left[\theta +\frac{(m-1)\pi }{4}\right]\mathrm{cosec}\left(\theta +\frac{m\pi }{4}\right)\\ =4\sqrt{2}\\ \Rightarrow {\sum }_{m=1}^6\frac{\sin \left[\theta +\frac{m\pi }{4}-\left(\theta +\frac{(m-1)\pi }{4}\right)\right]}{\sin \frac{\pi }{4}\left\{\sin \left(\theta +\frac{(m-1)\pi }{4}\right)\sin \left(\theta +\frac{m\pi }{4}\right)\right\}}\\ =4\sqrt{2}\end{array}$$ $$\begin{array}{r}\Rightarrow {\sum }_{m=1}^6\frac{\cot \left(\theta +\frac{(m-1)\pi }{4}\right)-\cot \left(\theta +\frac{m\pi }{4}\right)}{1/2}\\ =4\sqrt{2}\end{array}$$ $$\begin{array}{rl}& \Rightarrow \sum _{m=1}^6[\cot \left(\theta +\frac{(m-1)\pi }{4}\right)\\ & -\cot \left(\theta +\frac{m\pi }{4}\right)]=4\\ & \Rightarrow \cot (\theta )-\cot \left(\theta +\frac{\pi }{4}\right)+\cot \left(\theta +\frac{\pi }{4}\right)\\ & -\cot \left(\theta +\frac{2\pi }{4}\right)+\dots +\cot \left(\theta +\frac{5\pi }{4}\right)\\ & -\cot \left(\theta +\frac{6\pi }{4}\right)=4\\ & \Rightarrow \cot \theta -\cot \left(\frac{3\pi }{2}+\theta \right)=4\\ & \Rightarrow \cot \theta +\tan \theta =4\\ & \Rightarrow {\tan }^2\theta -4\tan \theta +1=0\\ & \Rightarrow (\tan \theta -2{)}^2-3=0\\ & \Rightarrow (\tan \theta -2+\sqrt{3})(\tan \theta -2-\sqrt{3})=0\\ & \Rightarrow \tan \theta =2-\sqrt{3}\text{ or }\\ & \tan \theta =2+\sqrt{3}\\ & \Rightarrow \theta =\frac{\pi }{12};\theta =\frac{5\pi }{12}\Rightarrow \theta \in \left(0,\frac{\pi }{2}\right)\\ & \end{array}$$