If $\operatorname{cosec}^2(\alpha+\beta)-\sin^2(\beta-\alpha)+\sin^2(2\alpha-\beta)=\cos^2(\alpha-\beta)$ where $\alpha, \beta \in (0, \frac{\pi}{2})$,then $\sin(\alpha-\beta)$ is equal to

If $x: y: z = \tan \left(\frac{\pi}{15}+\alpha\right): \tan \left(\frac{\pi}{15}+\beta\right): \tan \left(\frac{\pi}{15}+\gamma\right)$,then find the value of $\frac{z+x}{z-x} \sin ^2(\gamma-\alpha)+\frac{x+y}{x-y} \sin ^2(\alpha-\beta)+\frac{y+z}{y-z} \sin ^2(\beta-\gamma)$.

The value of the expression $\frac{(\sin 36^{\circ} + \cos 36^{\circ} - \sqrt{2} \sin 27^{\circ})^2}{2 \sin 54^{\circ}}$ is less than

Let the range of the function $f(x) = 6 + 16 \cos x \cdot \cos \left(\frac{\pi}{3} - x\right) \cdot \cos \left(\frac{\pi}{3} + x\right) \sin 3x \cdot \cos 6x$,where $x \in R$,be $[\alpha, \beta]$. Then the distance of the point $(\alpha, \beta)$ from the line $3x + 4y + 12 = 0$ is:

For non-negative integers $n$,let $f(n) = \frac{\sum_{k=0}^n \sin \left(\frac{k+1}{n+2} \pi\right) \sin \left(\frac{k+2}{n+2} \pi\right)}{\sum_{k=0}^n \sin ^2\left(\frac{k+1}{n+2} \pi\right)}$. Assuming $\cos ^{-1} x$ takes values in $[0, \pi]$,which of the following options is/are correct?
$(1)$ $\sin \left(7 \cos ^{-1} f(5)\right)=0$
$(2)$ $f(4)=\frac{\sqrt{3}}{2}$
$(3)$ $\lim _{n \rightarrow \infty} f(n)=\frac{1}{2}$
$(4)$ If $\alpha=\tan \left(\cos ^{-1} f(6)\right)$,then $\alpha^2+2 \alpha-1=0$

If $\tan \left(\frac{\pi}{4}+\alpha\right)=\tan ^3\left(\frac{\pi}{4}+\beta\right)$,then $\tan (\alpha+\beta) \cot (\alpha-\beta)=$