Let f(theta) = (1 + sin^2 theta)(2 - sin^2 th | vedclass

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(D) Let $x = \sin^2 	heta$. Since $0 \leq \sin^2 	heta \leq 1$, we have $0 \leq x \leq 1$. $f(	heta) = (1 + x)(2 - x) = 2 - x + 2x - x^2 = -x^2 + x

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$2 \leq f(	heta) \leq \frac{9}{4}$

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(D) Let $x = \sin^2 	heta$. Since $0 \leq \sin^2 	heta \leq 1$, we have $0 \leq x \leq 1$. $f(	heta) = (1 + x)(2 - x) = 2 - x + 2x - x^2 = -x^2 + x + 2$. To find the range, complete the square: $f(	heta) = -(x^2 - x) + 2 = -(x^2 - x + \frac{1}{4} - \frac{1}{4}) + 2 = -(x - \frac{1}{2})^2 + \frac{1}{4} + 2 = \frac{9}{4} - (x - \frac{1}{2})^2$. Since $0 \leq x \leq 1$, the term $(x - \frac{1}{2})^2$ ranges from $0$ (at $x = \frac{1}{2}$) to $\frac{1}{4}$ (at $x = 0$ or $x = 1$). Thus, the maximum value is $\frac{9}{4} - 0 = \frac{9}{4}$ and the minimum value is $\frac{9}{4} - \frac{1}{4} = 2$. Therefore, $2 \leq f(	heta) \leq \frac{9}{4}$.

Let $f(\theta) = (1 + \sin^2 \theta)(2 - \sin^2 \theta)$. Then, for all values of $\theta$:

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