<s>[INST] Show that $\cot\left(\frac{\pi}{4}+\beta\right)+\frac{1+\cot\beta}{1-\cot\beta}=-2\tan2\beta$ Show that $$\cot\left(\dfrac{\pi}{4}+\beta\right)+\dfrac{1+\cot\beta}{1-\cot\beta}=-2\tan2\beta$$
I'm supposed to solve this problem only with sum and difference formulas (identities).
So the LHS is $$\dfrac{\cot\dfrac{\pi}{4}\cot\beta-1}{\cot\dfrac{\pi}{4}+\cot\beta}+\dfrac{1+\cot\beta}{1-\cot\beta}=\dfrac{\cot\beta-1}{1+\cot\beta}+\dfrac{1+\cot\beta}{1-\cot\beta}=\dfrac{4\cot\beta}{1-\cot^2\beta}$$ I also tried to work with $\sin\beta$ and $\cos\beta$ and arrived at $$\dfrac{4\sin\beta\cos\beta}{\sin^2\beta-\cos^2\beta}$$ I don't see how to get $-2\tan2\beta$ from here (even with other identities).
 [/INST] $$\frac{4 \sin \beta \cos \beta}{\sin ^{2} \beta-\cos ^{2} \beta} = \frac{2 \sin2\beta}{-(\cos ^{2} \beta-\sin ^{2} \beta)}=-2\frac{\sin  2\beta}{\cos  2\beta}=-2 \tan(2\beta)$$
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