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7.5 I · Q62

Q.lim⁡x→π/2[cosec⁡x−1(π2−x)2]\displaystyle\lim_{x\to \pi/2}\left[\frac{\operatorname{cosec}x-1}{\left(\dfrac{\pi}{2}-x\right)^2}\right]

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✓ Free question

Let t=π/2−xt=\pi/2-x, so x=π/2−tx=\pi/2-t and t→0t\to0. Then sin⁡x=cos⁡t\sin x=\cos t, so csc⁡x=1/cos⁡t\csc x=1/\cos t, and csc⁡x−1=1−cos⁡tcos⁡t=2sin⁡2(t/2)cos⁡t\csc x-1=\dfrac{1-\cos t}{\cos t}=\dfrac{2\sin^2(t/2)}{\cos t}. The denominator is t2t^2. So the expression is 2sin⁡2(t/2)t2cos⁡t=12(sin⁡(t/2)t/2)21cos⁡t→12(1)211=12\dfrac{2\sin^2(t/2)}{t^2\cos t}=\dfrac12\left(\dfrac{\sin(t/2)}{t/2}\right)^2\dfrac{1}{\cos t}\to\dfrac12(1)^2\dfrac11=\dfrac12.

✓Final answer

12\dfrac{1}{2}

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