Both can't be compared.

A) Here, $f(n)$=$n^{(1-sin n)}$ and $g(n)=n^{(1+cos n)}$

when $n=0$ then $f(0)=g(0)=0$

when $n=\pi /4 $ then $f(\pi/4)>g(\pi /4)$

when $n=\pi /2 $ then $g(\pi/2)>f(\pi /2)$

when $n=\pi $ then $f(\pi)>g(\pi)$

when $n=1 $ then $f(1)=g(1)$.

Since, trigonometric functions are periodic in nature. both sin and cosine has period = $2 \pi.$ It means after $2 \pi$ they will show the same nature again and again for particular value of 'n'. someone will increase sometimes or someone will decrease sometimes because both sin and cosine functions oscillates between 0 and 1.

Now, to prove it asymptotically,

$\lim_{n\rightarrow \infty } \frac{f(n)}{g(n)} = \frac{1-sin n}{ 1+cosn}$ because base is same. Now limit does not exist here So, we can't say which has higher growth rate or we can say both can't be compared.

Same reasoning is for $(B)$

A) Here, $f(n)$=$n^{(1-sin n)}$ and $g(n)=n^{(1+cos n)}$

when $n=0$ then $f(0)=g(0)=0$

when $n=\pi /4 $ then $f(\pi/4)>g(\pi /4)$

when $n=\pi /2 $ then $g(\pi/2)>f(\pi /2)$

when $n=\pi $ then $f(\pi)>g(\pi)$

when $n=1 $ then $f(1)=g(1)$.

Since, trigonometric functions are periodic in nature. both sin and cosine has period = $2 \pi.$ It means after $2 \pi$ they will show the same nature again and again for particular value of 'n'. someone will increase sometimes or someone will decrease sometimes because both sin and cosine functions oscillates between 0 and 1.

Now, to prove it asymptotically,

$\lim_{n\rightarrow \infty } \frac{f(n)}{g(n)} = \frac{1-sin n}{ 1+cosn}$ because base is same. Now limit does not exist here So, we can't say which has higher growth rate or we can say both can't be compared.

Same reasoning is for $(B)$