 # If cosx= -12/13 and csc x<0, find cot(2x) floymdiT 2021-02-25 Answered

If $\mathrm{cos}x=-\frac{12}{13}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{csc}x<0$,
find $\mathrm{cot}\left(2x\right)$

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$\mathrm{cot}\left(2x\right)=\frac{\mathrm{cos}\left(2x\right)}{\mathrm{sin}\left(2x\right)}=\frac{1-2{\left(\mathrm{sin}\left(x\right)\right)}^{2}}{2}\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right).$
In the right-angled triangle ABC drawn on x-y axis, A is point (-12,0), B is (0,0) and C is (0,-5).
Angle ABC is a right-angle and BAC=x.
$\mathrm{cos}\left(x\right)=A\frac{B}{A}C=-\frac{12}{13}.BC=\sqrt{A{C}^{2}-A{B}^{2}}=\sqrt{{13}^{2}-{12}^{2}}=\sqrt{25}=5.$
$\mathrm{sin}\left(x\right)=B\frac{C}{A}C=-\frac{5}{13}\left(\mathrm{cos}ec\left(x\right)<0\right).\mathrm{cot}\left(2x\right)=\left(1-2\frac{\frac{25}{169}}{2\cdot \left(-\frac{5}{13}\right)\left(-\frac{12}{13}\right)=\frac{169-50}{120}=\frac{119}{120}}\right).$

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