1. Verificar la identidad: $\sin x \cdot \cot x \cdot \sec x = 1$ Usamos las definiciones: $\cot x = \frac{\cos x}{\sin x}$ y $\sec x = \frac{1}{\cos x}$. Entonces: $$\sin x \cdot \cot x \cdot \sec x = \sin x \cdot \frac{\cos x}{\sin x} \cdot \frac{1}{\cos x}$$ Cancelamos $\sin x$ y $\cos x$: $$= \cancel{\sin x} \cdot \frac{\cos x}{\cancel{\sin x}} \cdot \frac{1}{\cancel{\cos x}} = 1$$ 2. Verificar: $1 + \cot^2 x = \csc^2 x$ Sabemos que $\cot x = \frac{\cos x}{\sin x}$ y $\csc x = \frac{1}{\sin x}$. Entonces: $$1 + \cot^2 x = 1 + \left(\frac{\cos x}{\sin x}\right)^2 = 1 + \frac{\cos^2 x}{\sin^2 x} = \frac{\sin^2 x + \cos^2 x}{\sin^2 x} = \frac{1}{\sin^2 x} = \csc^2 x$$ 3. Verificar: $(\sin x - \cos x)(\csc x + \sec x) = \tan x - \cot x$ Expresamos: $\csc x = \frac{1}{\sin x}$, $\sec x = \frac{1}{\cos x}$, $\tan x = \frac{\sin x}{\cos x}$, $\cot x = \frac{\cos x}{\sin x}$. Multiplicamos: $$(\sin x - \cos x)\left(\frac{1}{\sin x} + \frac{1}{\cos x}\right) = (\sin x - \cos x)\frac{\cos x + \sin x}{\sin x \cos x}$$ Multiplicamos numerador: $$(\sin x - \cos x)(\cos x + \sin x) = \sin^2 x + \sin x \cos x - \cos x \sin x - \cos^2 x = \sin^2 x - \cos^2 x$$ Entonces: $$\frac{\sin^2 x - \cos^2 x}{\sin x \cos x}$$ Por otro lado: $$\tan x - \cot x = \frac{\sin x}{\cos x} - \frac{\cos x}{\sin x} = \frac{\sin^2 x - \cos^2 x}{\sin x \cos x}$$ Ambos lados son iguales. 4. Verificar: $\sin^2 x - \sin^2 x \cdot \cos^2 x = \sin^4 x$ Factorizamos $\sin^2 x$: $$\sin^2 x (1 - \cos^2 x) = \sin^2 x \cdot \sin^2 x = \sin^4 x$$ 5. Verificar: $(\sin x + \cos x)^2 - \tan x \cdot \cos^2 x = 1 + \sin x \cdot \cos x$ Expandimos: $$(\sin x + \cos x)^2 = \sin^2 x + 2 \sin x \cos x + \cos^2 x = 1 + 2 \sin x \cos x$$ Calculamos $\tan x \cdot \cos^2 x$: $$\tan x \cdot \cos^2 x = \frac{\sin x}{\cos x} \cdot \cos^2 x = \sin x \cos x$$ Entonces: $$1 + 2 \sin x \cos x - \sin x \cos x = 1 + \sin x \cos x$$ 6. Verificar: $(\tan x - \cot x)^2 = \tan^2 x + \cot^2 x - 2$ Expandimos: $$(\tan x - \cot x)^2 = \tan^2 x - 2 \tan x \cot x + \cot^2 x$$ Sabemos que $\tan x \cot x = 1$, entonces: $$= \tan^2 x + \cot^2 x - 2$$ 7. Verificar: $\tan x + \cot x = \sec x \cdot \csc x$ Expresamos: $$\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x}$$ Por otro lado: $$\sec x \cdot \csc x = \frac{1}{\cos x} \cdot \frac{1}{\sin x} = \frac{1}{\sin x \cos x}$$ Son iguales. 8. Verificar: $\frac{\sin^2 x}{1 + \cos x} = 1 - \cos x$ Multiplicamos ambos lados por $1 + \cos x$: $$\sin^2 x = (1 - \cos x)(1 + \cos x) = 1 - \cos^2 x$$ Sabemos que $\sin^2 x = 1 - \cos^2 x$, por lo que la identidad es correcta. 9. Verificar: $\frac{\sec x - 1}{\sin x} - \frac{1 - \cos x}{\tan x} = \tan x - \sin x$ Expresamos $\sec x = \frac{1}{\cos x}$ y $\tan x = \frac{\sin x}{\cos x}$: Primero: $$\frac{\frac{1}{\cos x} - 1}{\sin x} = \frac{\frac{1 - \cos x}{\cos x}}{\sin x} = \frac{1 - \cos x}{\sin x \cos x}$$ Segundo término: $$\frac{1 - \cos x}{\frac{\sin x}{\cos x}} = (1 - \cos x) \cdot \frac{\cos x}{\sin x} = \frac{(1 - \cos x) \cos x}{\sin x}$$ Entonces la expresión completa: $$\frac{1 - \cos x}{\sin x \cos x} - \frac{(1 - \cos x) \cos x}{\sin x} = \frac{1 - \cos x}{\sin x} \left(\frac{1}{\cos x} - \cos x\right) = \frac{1 - \cos x}{\sin x} \cdot \frac{1 - \cos^2 x}{\cos x}$$ Sabemos que $1 - \cos^2 x = \sin^2 x$, entonces: $$= \frac{1 - \cos x}{\sin x} \cdot \frac{\sin^2 x}{\cos x} = \frac{(1 - \cos x) \sin^2 x}{\sin x \cos x} = \frac{(1 - \cos x) \sin x}{\cos x}$$ Por otro lado: $$\tan x - \sin x = \frac{\sin x}{\cos x} - \sin x = \sin x \left(\frac{1}{\cos x} - 1\right) = \sin x \frac{1 - \cos x}{\cos x}$$ Ambos lados coinciden. 10. Verificar: $\cos x (\tan x + 1) - \cos x = \sin x$ Expandimos: $$\cos x \tan x + \cos x - \cos x = \cos x \tan x = \cos x \cdot \frac{\sin x}{\cos x} = \sin x$$ 11. Verificar: $\csc x - \sin x = \cot x \cdot \cos x$ Expresamos: $$\csc x - \sin x = \frac{1}{\sin x} - \sin x = \frac{1 - \sin^2 x}{\sin x} = \frac{\cos^2 x}{\sin x}$$ Por otro lado: $$\cot x \cdot \cos x = \frac{\cos x}{\sin x} \cdot \cos x = \frac{\cos^2 x}{\sin x}$$ Son iguales. 12. Verificar: $\tan x \cdot \cot x - \cos^2 x = \sin^2 x$ Sabemos que $\tan x \cdot \cot x = 1$, entonces: $$1 - \cos^2 x = \sin^2 x$$ 13. Verificar: $\frac{1}{\sin x \cdot \cos x} - \frac{\cos x}{\sin x} = \tan x$ Simplificamos: $$\frac{1}{\sin x \cos x} - \frac{\cos x}{\sin x} = \frac{1 - \cos^2 x}{\sin x \cos x} = \frac{\sin^2 x}{\sin x \cos x} = \frac{\sin x}{\cos x} = \tan x$$ 14. Verificar: $\frac{\sin x + \cos x}{\csc x + \sec x} = \frac{\sin x}{\sec x}$ Expresamos: $$\csc x + \sec x = \frac{1}{\sin x} + \frac{1}{\cos x} = \frac{\cos x + \sin x}{\sin x \cos x}$$ Entonces: $$\frac{\sin x + \cos x}{\frac{\cos x + \sin x}{\sin x \cos x}} = (\sin x + \cos x) \cdot \frac{\sin x \cos x}{\cos x + \sin x} = \sin x \cos x$$ Por otro lado: $$\frac{\sin x}{\sec x} = \sin x \cdot \cos x = \sin x \cos x$$ Son iguales. 15. Verificar: $\frac{\sec x + \tan x}{\cos x + \cot x} = \sec x \cdot \tan x$ Expresamos: $$\sec x + \tan x = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{1 + \sin x}{\cos x}$$ $$\cos x + \cot x = \cos x + \frac{\cos x}{\sin x} = \frac{\cos x \sin x + \cos x}{\sin x} = \frac{\cos x (\sin x + 1)}{\sin x}$$ Entonces: $$\frac{\frac{1 + \sin x}{\cos x}}{\frac{\cos x (1 + \sin x)}{\sin x}} = \frac{1 + \sin x}{\cos x} \cdot \frac{\sin x}{\cos x (1 + \sin x)} = \frac{\sin x}{\cos^2 x} = \sec x \tan x$$ 16. Verificar: $(\sec^2 x - 1) \cdot \cos \sec^2 x \cdot \cot x = \frac{1}{\cos x \sin x}$ Sabemos que $\sec^2 x - 1 = \tan^2 x$. Entonces: $$(\sec^2 x - 1) \cdot \cos x \cdot \sec^2 x \cdot \cot x = \tan^2 x \cdot \cos x \cdot \sec^2 x \cdot \cot x$$ Expresamos: $$\tan^2 x = \frac{\sin^2 x}{\cos^2 x}, \sec^2 x = \frac{1}{\cos^2 x}, \cot x = \frac{\cos x}{\sin x}$$ Multiplicamos: $$\frac{\sin^2 x}{\cos^2 x} \cdot \cos x \cdot \frac{1}{\cos^2 x} \cdot \frac{\cos x}{\sin x} = \frac{\sin^2 x}{\cos^2 x} \cdot \frac{\cos x \cos x}{\cos^2 x} \cdot \frac{1}{\sin x} = \frac{\sin^2 x}{\cos^2 x} \cdot \frac{\cos^2 x}{\cos^2 x} \cdot \frac{1}{\sin x} = \frac{\sin^2 x}{\cos^2 x} \cdot \frac{1}{\sin x}$$ Simplificamos: $$= \frac{\sin x}{\cos^2 x}$$ Pero la expresión original es: $$\frac{1}{\cos x \sin x}$$ Hay un error en la expresión dada o en la interpretación, ya que no coinciden. 17. Verificar: $\frac{\sin x \cdot \sec x}{\tan x + \cot x} = 1 - \cos^2 x$ Expresamos: $$\sec x = \frac{1}{\cos x}, \tan x = \frac{\sin x}{\cos x}, \cot x = \frac{\cos x}{\sin x}$$ Entonces: $$\frac{\sin x \cdot \frac{1}{\cos x}}{\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}} = \frac{\frac{\sin x}{\cos x}}{\frac{\sin^2 x + \cos^2 x}{\sin x \cos x}} = \frac{\frac{\sin x}{\cos x}}{\frac{1}{\sin x \cos x}} = \frac{\sin x}{\cos x} \cdot \sin x \cos x = \sin^2 x$$ Sabemos que $1 - \cos^2 x = \sin^2 x$, por lo que la identidad es correcta. 18. Verificar: $\frac{1}{\cos x} - \frac{\cos x}{1 + \sin x} = \tan x$ Multiplicamos por $1 + \sin x$: $$\frac{1 + \sin x}{\cos x} - \frac{\cos^2 x}{1 + \sin x} = \tan x (1 + \sin x)$$ Pero mejor simplificamos directamente: $$\frac{1}{\cos x} - \frac{\cos x}{1 + \sin x} = \frac{1 + \sin x}{\cos x (1 + \sin x)} - \frac{\cos^2 x}{\cos x (1 + \sin x)} = \frac{1 + \sin x - \cos^2 x}{\cos x (1 + \sin x)}$$ Sabemos que $1 - \cos^2 x = \sin^2 x$, entonces: $$= \frac{\sin^2 x + \sin x}{\cos x (1 + \sin x)} = \frac{\sin x (\sin x + 1)}{\cos x (1 + \sin x)} = \frac{\sin x}{\cos x} = \tan x$$ 19. Verificar: $\frac{1 - \tan^2 x}{1 + \tan^2 x} = 1 - 2 \sin^2 x$ Sabemos que $\tan^2 x = \frac{\sin^2 x}{\cos^2 x}$. Entonces: $$\frac{1 - \frac{\sin^2 x}{\cos^2 x}}{1 + \frac{\sin^2 x}{\cos^2 x}} = \frac{\frac{\cos^2 x - \sin^2 x}{\cos^2 x}}{\frac{\cos^2 x + \sin^2 x}{\cos^2 x}} = \frac{\cos^2 x - \sin^2 x}{\cos^2 x + \sin^2 x} = \cos 2x$$ Sabemos que $\cos 2x = 1 - 2 \sin^2 x$, por lo que la identidad es correcta. Finalizamos la verificación de las 19 identidades trigonométricas.