If a+bcsc⁡x=a−bcot⁡x show that csc⁡xcot⁡x=a2−b24ab

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Suppose sin⁡x≠0. Then we have that csc⁡x=1sin⁡x and cos⁡xsin⁡x=cot⁡x  This implies in cos⁡x=a−ba+b. Therefore, csc⁡x⋅cot⁡x=cos⁡xsin2x=cos⁡x1−cos2x  Notice that 1−cos2x=1−(a−ba+b)2=4ab(a+b)2  Then, cos⁡x1−cos2x=(a−b)(a+b)4ab=a2−b24ab

If a+bcsc⁡x=a−bcot⁡x show that csc⁡xcot⁡x=a2−b24ab

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