Find limtan^(2)(theta)[1-sin(theta)]. theta rarr(pi)/(2) Choose 1 answer: (A) 0 (B) (1)/(2) (C) -2 (D) The limit doesn't exist

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Find  limtan^(2)(theta)[1-sin(theta)].  theta rarr(pi)/(2) Choose 1 answer: (A) 0 (B)  (1)/(2) (C) -2 (D) The limit doesn't exist

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Identify Behavior: Identify the behavior of the function as θ approaches π/2. As θ approaches π/2, sin(θ) approaches 1. This means that the term (1 - sin(θ)) approaches 0. However, tan(θ) approaches infinity as θ approaches π/2 from the left, and negative infinity from the right, because tan(θ) = sin(θ)/cos(θ) and cos(θ) approaches 0. The square of tan(θ), which is tan^2(θ), will approach positive infinity from both sides since squaring a negative number results in a positive number. We have an indeterminate form of type [∞ * 0], which means we need to manipulate the expression to resolve the indeterminate form.Apply L'Hôpital's Rule: Apply L'Hôpital's Rule to resolve the indeterminate form. L'Hôpital's Rule states that if the limit of f(θ)/g(θ) as θ approaches a value c is of the form 0/0 or ∞/∞, then the limit of f(θ)/g(θ) as θ approaches c is the same as the limit of f'(θ)/g'(θ) as θ approaches c, provided that the derivatives exist and the limit of f'(θ)/g'(θ) is determinate. To apply L'Hôpital's Rule, we need to rewrite the expression in a form that allows us to use the rule. We can rewrite tan^2(θ)(1 - sin(θ)) as [tan(θ)]^2 * (1 - sin(θ)) and then as [sin(θ)/cos(θ)]^2 * (1 - sin(θ)). To use L'Hôpital's Rule, we need a fraction, so we can set f(θ) = [sin(θ)]^2 * (1 - sin(θ)) and g(θ) = cos^2(θ). Now we have f(θ)/g(θ) = [sin(θ)]^2 * (1 - sin(θ)) / cos^2(θ).Differentiate Functions: Differentiate f(θ) and g(θ) with respect to θ. The derivative of f(θ) = [sin(θ)]^2 * (1 - sin(θ)) is f'(θ) = 2sin(θ)cos(θ) * (1 - sin(θ)) - [sin(θ)]^2 * cos(θ) using the product rule and chain rule. The derivative of g(θ) = cos^2(θ) is g'(θ) = -2cos(θ)sin(θ) using the chain rule.Apply L'Hôpital's Rule Correction: Apply L'Hôpital's Rule by taking the limit of f'(θ)/g'(θ) as θ approaches π/2. We now take the limit of f'(θ)/g'(θ) as θ approaches π/2. However, there is a mistake in the differentiation process in the previous step. The correct derivative of f(θ) should be f'(θ) = 2sin(θ)cos(θ) * (1 - sin(θ)) - [sin(θ)]^2 * cos(θ) = 2sin(θ)cos(θ) - 2sin^2(θ)cos(θ) - [sin(θ)]^2 * cos(θ). We need to correct this before proceeding.

Find $\lim_{\theta \rightarrow \frac{\pi}{2}} \tan ^{2}(\theta)[1-\sin (\theta)]$ . $\newline$ Choose $1$ answer: $\newline$ (A) $0$ $\newline$ (B) $\frac{1}{2}$ $\newline$ (C) $-2$ $\newline$ (D) The limit doesn't exist

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Find $\lim_{\theta \rightarrow \frac{\pi}{2}} \tan ^{2}(\theta)[1-\sin (\theta)]$ . $\newline$ Choose $1$ answer: $\newline$ (A) $0$ $\newline$ (B) $\frac{1}{2}$ $\newline$ (C) $-2$ $\newline$ (D) The limit doesn't exist