If A = sin^8 theta + cos^{14} theta,then for | vedclass

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(B) Given that $A = \sin^8 	heta + \cos^{14} 	heta$.Since $0 \le \sin^2 	heta \le 1$ and $0 \le \cos^2 	heta \le 1$ for all real $	heta$,we have:

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$0 < A \le 1$

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(B) Given that $A = \sin^8 	heta + \cos^{14} 	heta$.Since $0 \le \sin^2 	heta \le 1$ and $0 \le \cos^2 	heta \le 1$ for all real $	heta$,we have:$\sin^8 	heta \le \sin^2 	heta$ and $\cos^{14} 	heta \le \cos^2 	heta$.Adding these inequalities,we get:$A = \sin^8 	heta + \cos^{14} 	heta \le \sin^2 	heta + \cos^2 	heta = 1$.Also,since $\sin^8 	heta > 0$ and $\cos^{14} 	heta > 0$ for all $	heta$ (excluding cases where one is zero,but they cannot both be zero simultaneously),$A > 0$.Therefore,$0 < A \le 1$.

If $A = \sin^8 \theta + \cos^{14} \theta$,then for all real values of $\theta$:

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