If f:[0, pi / 2) rightarrow R is defined as f | vedclass

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(C) Given the determinant $f(	heta) = \left|\begin{array}{ccc} 1 & 	an 	heta & 1 \ -	an 	heta & 1 & 	an 	heta \ -1 & -	an 	heta & 1 \end{ar

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$[2, \infty)$

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(C) Given the determinant $f(	heta) = \left|\begin{array}{ccc} 1 & 	an 	heta & 1 \ -	an 	heta & 1 & 	an 	heta \ -1 & -	an 	heta & 1 \end{array}ight|$.Expanding along the first row:$f(	heta) = 1(1 - (-	an^2 	heta)) - 	an 	heta(-	an 	heta - (-	an 	heta)) + 1(	an^2 	heta - (-1))$$f(	heta) = 1(1 + 	an^2 	heta) - 	an 	heta(0) + 1(	an^2 	heta + 1)$$f(	heta) = (1 + 	an^2 	heta) + (1 + 	an^2 	heta) = 2(1 + 	an^2 	heta) = 2 \sec^2 	heta$.Since $	heta \in [0, \pi/2)$, $	an 	heta \in [0, \infty)$, so $\sec^2 	heta \in [1, \infty)$.Therefore, $f(	heta) = 2 \sec^2 	heta \in [2, \infty)$.

If $f:[0, \pi / 2) \rightarrow R$ is defined as $f(\theta)=\left|\begin{array}{ccc}1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{array}\right|$, then the range of $f$ is:

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