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Solution

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Solution

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Solution

Maximum of $\cot A\cot B\cot C$ occurs at $A=B=C=\dfrac{\pi}{3}$ $K_{\max}=\cot^3\dfrac{\pi}{3}=\left(\dfrac{1}{\sqrt3}\right)^3=\dfrac{1}{3\sqrt3}$

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Solution

Given $\displaystyle \sum_{m=1}^{6} \csc!\left(\theta+\dfrac{(m-1)\pi}{4}\right), \csc!\left(\theta+\dfrac{m\pi}{4}\right)=4\sqrt{2}$ Use the identity $\csc x \csc y=\dfrac{\cot x-\cot y}{\sin(y-x)}$ Here, $y-x=\dfrac{\pi}{4}$ and $\sin\dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}$ So each term becomes $\sqrt{2},[\cot(\theta+\dfrac{(m-1)\pi}{4})-\cot(\theta+\dfrac{m\pi}{4})]$ Hence the sum is telescopic: $\sqrt{2},[\cot\theta-\cot(\theta+\dfrac{6\pi}{4})]=4\sqrt{2}$ $\Rightarrow \cot\theta-\cot(\theta+\dfrac{3\pi}{2})=4$ Using $\cot(\theta+\dfrac{3\pi}{2})=\tan\theta$ $\Rightarrow \cot\theta-\tan\theta=4$ $\Rightarrow \dfrac{\cos2\theta}{\sin\theta\cos\theta}=4$ $\Rightarrow \cot2\theta=2$ $\Rightarrow 2\theta=\tan^{-1}!\left(\dfrac{1}{2}\right)$ $\Rightarrow \theta=\dfrac{\pi}{12},\ \dfrac{5\pi}{12}$

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Solution

Since $\sin\beta$ is the G.M. of $\sin\alpha$ and $\cos\alpha$, $\sin^2\beta=\sin\alpha\cos\alpha=\dfrac12\sin2\alpha$ So, $\cos2\beta=1-2\sin^2\beta=1-\sin2\alpha$ But $1-\sin2\alpha=2\sin^2\left(\dfrac{\pi}{4}-\alpha\right)$