$$\sin^2\alpha+\cos^2\alpha=1;\;\;\;\text{tg}\alpha\cdot\text{ctg}\alpha=1;$$
$$\text{tg}\alpha=\frac{1}{\text{ctg}\alpha};\;\;\;\text{ctg}\alpha=\frac{1}{\text{tg}\alpha};$$
$$\text{tg}\alpha=\frac{\sin\alpha}{\cos\alpha};\;\;\;\text{ctg}\alpha=\frac{\cos\alpha}{\sin\alpha};$$
$$1+\text{tg}^2\alpha=\frac{1}{\cos^2\alpha};\;\;\;1+\text{ctg}^2\alpha=\frac{1}{\sin^2\alpha};$$
$$|\sin\alpha|=\sqrt{1-\cos^2\alpha}=\frac{|\text{tg}\alpha|}{\sqrt{1+\text{tg}^2\alpha}}=\frac{1}{\sqrt{1+\text{ctg}^2\alpha}};$$
$$|\cos\alpha|=\sqrt{1-\sin^2\alpha}=\frac{1}{\sqrt{1+\text{tg}^2\alpha}}=\frac{|\text{ctg}\alpha|}{\sqrt{1+\text{ctg}^2\alpha}};$$
$$|\text{tg}\alpha|=\frac{|\sin\alpha|}{\sqrt{1-\sin^2\alpha}}=\frac{\sqrt{1-\cos^2\alpha}}{|\cos\alpha|}=\frac{1}{|\text{ctg}\alpha|};$$
$$|\text{ctg}\alpha|=\frac{\sqrt{1-\sin^2\alpha}}{|\sin\alpha|}=\frac{|\cos\alpha|}{\sqrt{1-\cos^2\alpha}}=\frac{1}{|\text{tg}\alpha|}.$$