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I need some help with solving this Identity. I've tried several methods, but I haven't been able to come up with the correct answer and/or method.

I've tried several things. Ranging from converting the cot trigonometric function that's within the parentheses to Csc, and the same with the Csc function to the Cot (opting out the minus sign of course) to even changing them to sin and cos (which is silly in my opinion, but I was desperate to find the answer).

I can bet my money that the answer is as simple as can be. It's been awhile since I've had a math course, and I'm unaware of what I can do to manipulate the math to fit the needs of the answer.

Please help me from going insane.


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You mention converting the $\cot$ to $\csc$ ---presumably via the identity $\cot^2 + 1 = \csc^2$--- but perhaps you got off track.

$$\begin{align} 3\left(\cot^2\theta + 1 \right) - \csc^2\theta - 1 &= 3\csc^2\theta - \csc^2\theta - 1 \\ &=2\csc^2\theta - 1 \\[6pt] &=\csc^2\theta + \left( \csc^2\theta - 1 \right) \\[6pt] &=\csc^2\theta + \cot^2\theta \end{align}$$

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Lets, not make extra work for yourself. You already have a $\cot^2\theta$, so keep it. $$\begin{array}{lll} 3(\cot^2\theta+1)-\csc^2\theta-1&=&(\cot^2\theta+1)+2(\cot^2\theta+1)-\csc^2\theta-1\\ &=&\cot^2\theta+2(\cot^2\theta+1)-\csc^2\theta\\ &=&\cot^2\theta+2\csc^2\theta-\csc^2\theta\\ &=&\cot^2\theta+\csc^2\theta \end{array}$$

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