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Miscellaneous Exercise 3 (II) · Q104

Q.Prove the following: cosec 48°+cosec 96°+cosec 192°+cosec 384°=0\text{cosec}\,48°+\text{cosec}\,96°+\text{cosec}\,192°+\text{cosec}\,384°=0

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Step 1: Use the identity cosec θ=cot⁡θ2−cot⁡θ\text{cosec}\,\theta=\cot\dfrac{\theta}{2}-\cot\theta (from 1−2sin⁡2θ21-2\sin^2\frac\theta2 type manipulation, equivalently cot⁡θ2−cot⁡θ=cos⁡θ/2sin⁡θ/2−cos⁡θsin⁡θ=1sin⁡θ\cot\frac\theta2-\cot\theta=\frac{\cos\theta/2}{\sin\theta/2}-\frac{\cos\theta}{\sin\theta}=\frac{1}{\sin\theta}).

Step 2: cosec 48°=cot⁡24°−cot⁡48°\text{cosec}\,48°=\cot24°-\cot48°, cosec 96°=cot⁡48°−cot⁡96°\text{cosec}\,96°=\cot48°-\cot96°, cosec 192°=cot⁡96°−cot⁡192°\text{cosec}\,192°=\cot96°-\cot192°, cosec 384°=cot⁡192°−cot⁡384°\text{cosec}\,384°=\cot192°-\cot384°.

Step 3: Adding all four, every middle term cancels (telescoping): the sum =cot⁡24°−cot⁡384°=\cot24°-\cot384°. …

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