COMEDK · Maths · 16. Limits
\(\lim _{\theta \rightarrow \dfrac{\pi}{2}} \dfrac{1-\sin \theta}{\left(\dfrac{\pi}{2}-\theta\right) \cos \theta}\) is equal to
- A \(-1\)
- B \(-\dfrac{1}{2}\)
- C \(1\)
- D \(\dfrac{1}{2}\)
Answer & Solution
Correct Answer
(D) \(\dfrac{1}{2}\)
Step-by-step Solution
Detailed explanation
Let \(x = \dfrac{\pi}{2} - \theta\). As \(\theta \rightarrow \dfrac{\pi}{2}\), \(x \rightarrow 0\). Substituting \(\theta = \dfrac{\pi}{2} - x\), we have \(\sin \theta = \sin(\dfrac{\pi}{2} - x) = \cos x\) and \(\cos \theta = \cos(\dfrac{\pi}{2} - x) = \sin x\). The limit…
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