If cosx= -12/13 and csc x<0, find cot(2x)

Question
If \(\displaystyle{\cos{{x}}}=-\frac{{12}}{{13}}{\quad\text{and}\quad}{\csc{{x}}}{<}{0}\)</span>,
find \(\displaystyle{\cot{{\left({2}{x}\right)}}}\)

Answers (1)

2021-02-26
\(\displaystyle{\cot{{\left({2}{x}\right)}}}=\frac{{\cos{{\left({2}{x}\right)}}}}{{\sin{{\left({2}{x}\right)}}}}=\frac{{{1}-{2}{\left({\sin{{\left({x}\right)}}}\right)}^{{2}}}}{{2}}{\sin{{\left({x}\right)}}}{\cos{{\left({x}\right)}}}.\)
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.
\(\displaystyle{\cos{{\left({x}\right)}}}={A}\frac{{B}}{{A}}{C}=-\frac{{12}}{{13}}.{B}{C}=\sqrt{{{A}{C}^{{2}}-{A}{B}^{{2}}}}=\sqrt{{{13}^{{2}}-{12}^{{2}}}}=\sqrt{{{25}}}={5}.\)
\(\displaystyle{\sin{{\left({x}\right)}}}={B}\frac{{C}}{{A}}{C}=-\frac{{5}}{{13}}{\left({\cos{{e}}}{c}{\left({x}\right)}{<}{0}\right)}.{\cot{{\left({2}{x}\right)}}}={\left({1}-{2}\frac{{\frac{{25}}{{169}}}}{{{2}\cdot{\left(-\frac{{5}}{{13}}\right)}{\left(-\frac{{12}}{{13}}\right)}=\frac{{{169}-{50}}}{{120}}=\frac{{119}}{{120}}.}}\right.}\)</span>
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