If \(\displaystyle{\tan{{\left({x}\right)}}}=\frac{{3}}{{4}}\), then \(\displaystyle{\sin{{\left({x}\right)}}}=\frac{{3}}{{5}}{\quad\text{and}\quad}{\cos{{\left({x}\right)}}}=\frac{{4}}{{5}}\), because of Pythagoras’ Theorem.

This is the 3-4-5 right triangle. \(\displaystyle{\tan{{\left({2}{x}\right)}}}={2}\frac{{\tan{{\left({x}\right)}}}}{{{1}-{{\tan}^{{2}}{\left({x}\right)}}}}=\frac{{\frac{{3}}{{2}}}}{{{1}-\frac{{9}}{{16}}}}=\frac{{24}}{{7}}\).

If the legs of a right triangle are 24 and 7, the hypotenuse is \(\displaystyle\sqrt{{{24}^{{2}}+{7}^{{2}}}}=\sqrt{{{576}+{49}}}=\sqrt{{625}}={25}\).

So \(\displaystyle{\sin{{\left({2}{x}\right)}}}=\frac{{24}}{{25}}\) and \(\displaystyle{\cos{{\left({2}{x}\right)}}}=\frac{{7}}{{25}}.\)

\(\displaystyle{\cot{{\left({2}{x}\right)}}}=\frac{{7}}{{24}},{\csc{{\left({2}{x}\right)}}}=\frac{{25}}{{24}},{\sec{{\left({2}{x}\right)}}}=\frac{{25}}{{7}}.\)

This is the 3-4-5 right triangle. \(\displaystyle{\tan{{\left({2}{x}\right)}}}={2}\frac{{\tan{{\left({x}\right)}}}}{{{1}-{{\tan}^{{2}}{\left({x}\right)}}}}=\frac{{\frac{{3}}{{2}}}}{{{1}-\frac{{9}}{{16}}}}=\frac{{24}}{{7}}\).

If the legs of a right triangle are 24 and 7, the hypotenuse is \(\displaystyle\sqrt{{{24}^{{2}}+{7}^{{2}}}}=\sqrt{{{576}+{49}}}=\sqrt{{625}}={25}\).

So \(\displaystyle{\sin{{\left({2}{x}\right)}}}=\frac{{24}}{{25}}\) and \(\displaystyle{\cos{{\left({2}{x}\right)}}}=\frac{{7}}{{25}}.\)

\(\displaystyle{\cot{{\left({2}{x}\right)}}}=\frac{{7}}{{24}},{\csc{{\left({2}{x}\right)}}}=\frac{{25}}{{24}},{\sec{{\left({2}{x}\right)}}}=\frac{{25}}{{7}}.\)