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Find 
lim_(theta rarr(pi)/(2))(sin^(2)(2theta))/(1-sin^(2)(theta))
Choose 1 answer:
(A) 1
(B) 2
(C) 4
(D) The limit doesn't exist

Find limθπ2sin2(2θ)1sin2(θ) \lim _{\theta \rightarrow \frac{\pi}{2}} \frac{\sin ^{2}(2 \theta)}{1-\sin ^{2}(\theta)} \newlineChoose 11 answer:\newline(A) 11\newline(B) 22\newline(C) 44\newline(D) The limit doesn't exist

Full solution

Q. Find limθπ2sin2(2θ)1sin2(θ) \lim _{\theta \rightarrow \frac{\pi}{2}} \frac{\sin ^{2}(2 \theta)}{1-\sin ^{2}(\theta)} \newlineChoose 11 answer:\newline(A) 11\newline(B) 22\newline(C) 44\newline(D) The limit doesn't exist
  1. Recognize Limit Issue: First, let's recognize that as θ\theta approaches π2\frac{\pi}{2}, sin(θ)\sin(\theta) approaches 11. This means that 1sin2(θ)1 - \sin^2(\theta) approaches 00, which could lead to a division by zero issue. We need to find a way to simplify the expression to avoid this.
  2. Trigonometric Identity: We can use the trigonometric identity sin2(x)=1cos2(x)\sin^2(x) = 1 - \cos^2(x) to rewrite the denominator. This gives us:\newlinelimθπ2sin2(2θ)1sin2(θ)=limθπ2sin2(2θ)cos2(θ)\lim_{\theta \rightarrow \frac{\pi}{2}}\frac{\sin^2(2\theta)}{1-\sin^2(\theta)} = \lim_{\theta \rightarrow \frac{\pi}{2}}\frac{\sin^2(2\theta)}{\cos^2(\theta)}
  3. Double Angle Formula: Next, we can use the double angle formula for sine, which is sin(2x)=2sin(x)cos(x)\sin(2x) = 2\sin(x)\cos(x). Applying this to sin2(2θ)\sin^2(2\theta), we get:\newlinesin2(2θ)=(2sin(θ)cos(θ))2=4sin2(θ)cos2(θ)\sin^2(2\theta) = (2\sin(\theta)\cos(\theta))^2 = 4\sin^2(\theta)\cos^2(\theta)
  4. Substitute and Simplify: Substituting this back into our limit expression, we have:\newlinelimθπ24sin2(θ)cos2(θ)cos2(θ)\lim_{\theta \rightarrow \frac{\pi}{2}}\frac{4\sin^2(\theta)\cos^2(\theta)}{\cos^2(\theta)}
  5. Cancel Terms: We can now cancel out the cos2(θ)\cos^2(\theta) terms in the numerator and the denominator, which simplifies our expression to:\newlinelimθπ24sin2(θ)\lim_{\theta \rightarrow \frac{\pi}{2}} 4\sin^2(\theta)
  6. Evaluate Limit: As θ\theta approaches π2\frac{\pi}{2}, sin(θ)\sin(\theta) approaches 11. Therefore, sin2(θ)\sin^2(\theta) approaches 121^2, which is 11. So our limit becomes:\newlinelimθπ24sin2(θ)=4×1=4\lim_{\theta \to \frac{\pi}{2}} 4\sin^2(\theta) = 4 \times 1 = 4
  7. Final Answer: The final answer is that the limit of sin2(2θ)/(1sin2(θ))\sin^2(2\theta)/(1-\sin^2(\theta)) as θ\theta approaches π/2\pi/2 is 44. Therefore, the correct choice is (C)(C) 44.

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