Match each entry in List - I to the correct entries in List - II

	LIST-I		LIST-II
I)	If  $\frac{{\sin }^{4}x}{2}+\frac{{\cos }^{4}x}{3}=\frac{1}{5}$  then ${\tan }^{2}x=$	P)	$4/5$
II)	$f_k\left(x\right)=\frac{1}{k}({\sin }^{k}x+{\cos }^{k}x)\text{ where }x\in R \text{ and k}\ge 1\text{ then }{f}_4\left(x\right)-f_6\left(x\right)=$	Q)	$1/6$
III)	${\sec }^4\theta +{\sec }^2\theta =10+{\tan }^4\theta +{\tan }^2\theta \text{ then }\mathrm{si}{\text{n}}^{\text{2}}\theta =$	R)	$2/3$
IV)	$5\tan \theta =4\text{ then }\frac{5\sin \theta -3\cos \theta }{5\sin \theta +2\cos \theta }\text{=}$	S)	$1/12$

I.  $\frac{{\sin }^{4}x}{2}+\frac{{\cos }^{4}x}{3}=\frac{1}{5}$
 $25{\sin }^{4}x-20{\sin }^{2}x+4=0$
 ${\sin }^{2}x=\frac{2}{5}{\cos }^{2}x=\frac{3}{5}$
 ${\tan }^{2}x=\frac{2}{3}$
II.  $f_4\left(x\right)+f_6\left(x\right)$
 $\frac{3({\sin }^{4}x+{\cos }^{4}x)-2({\sin }^{6}x+{\cos }^{6}x)}{12}$
  $\frac{3(1-2{\sin }^{2}x{\cos }^{2}x)-2(1-3{\sin }^{2}x{\cos }^{2}x)}{12}=\frac{1}{12}$
III.  ${\sec }^4\theta +{\sec }^2\theta =10+{\tan }^4\theta +{\tan }^2\theta$
  $1+{\tan }^4\theta +2{\tan }^2\theta +1+{\tan }^2\theta =10+{\tan }^4\theta +{\tan }^2\theta$
 $2+3{\tan }^2\theta =10+{\tan }^2\theta$
 $2{\tan }^2\theta =8{\tan }^2\theta =4{\sin }^2\theta =\frac{4}{5}$
IV.  $5\tan \theta =4\frac{5\tan \theta -3}{5\tan \theta +2}=\frac{5\times \frac{4}{5}-3}{5\times \frac{4}{5}+2}$