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Exercise 12.1 · Q21

Q.lim⁡x→0(cosec⁡x−cot⁡x)\lim_{x\to 0}(\operatorname{cosec} x - \cot x)

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As x→0x\to 0 this is the indeterminate form ∞−∞\infty-\infty. Combining over a common denominator gives 1−cos⁡xsin⁡x=tan⁡x2\dfrac{1-\cos x}{\sin x}=\tan\dfrac{x}{2}, whose limit is 00.

Near x=0x=0 both cosec⁡x=1sin⁡x\operatorname{cosec} x=\dfrac{1}{\sin x} and cot⁡x=cos⁡xsin⁡x\cot x=\dfrac{\cos x}{\sin x} blow up, so cosec⁡x−cot⁡x\operatorname{cosec} x-\cot x is the indeterminate form ∞−∞\infty-\infty. We first combine the two terms into a single fraction.

Step 1 — Write in terms of sin⁡x\sin x and cos⁡x\cos x.

cosec⁡x−cot⁡x=1sin⁡x−cos⁡xsin⁡x=1−cos⁡xsin⁡x.\operatorname{cosec} x-\cot x=\frac{1}{\sin x}-\frac{\cos x}{\sin x}=\frac{1-\cos x}{\sin x}.

Step 2 — Use the half-angle identities.

Recall 1−cos⁡x=2sin⁡2x21-\cos x=2\sin^2\dfrac{x}{2} and sin⁡x=2sin⁡x2cos⁡x2\sin x=2\sin\dfrac{x}{2}\cos\dfrac{x}{2}. Substituting, …

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