If $f(x) = \frac{\cos^2 x + \sin^4 x}{\sin^2 x + \cos^4 x}$ for all $x \in R$,then $f(2023) = $

If $\frac{\sqrt{2} \sin \alpha}{\sqrt{1+\cos 2 \alpha}}=\frac{1}{7}$ and $\sqrt{\frac{1-\cos 2 \beta}{2}}=\frac{1}{\sqrt{10}}$ where $\alpha, \beta \in (0, \frac{\pi}{2})$,then $\tan (\alpha+2 \beta)$ is equal to

If $\frac{\sin^4 x}{2} + \frac{\cos^4 x}{3} = \frac{1}{5}$,then $27 \sec^6 x + 8 \operatorname{cosec}^6 x = $

The value of the expression $\frac{1+\sin 2 \alpha}{\cos (2 \alpha-2 \pi) \tan \left(\alpha-\frac{3 \pi}{4}\right)} - \frac{1}{4} \sin 2 \alpha \left[\cot \frac{\alpha}{2}+\cot \left(\frac{3 \pi}{2}+\frac{\alpha}{2}\right)\right]$ is:

If $P = \sin \frac{2 \pi}{7} + \sin \frac{4 \pi}{7} + \sin \frac{8 \pi}{7}$ and $Q = \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{8 \pi}{7}$,then the point $(P, Q)$ lies on the circle of radius

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 $\alpha^2+2 \alpha-1=0$