d. \( (1+\csc \alpha) \cdot(1-\operatorname{sen} \alpha)=\operatorname{ctg} \alpha \cdot \csc \alpha \) b. \( \operatorname{tg}^{4} \alpha-\sec ^{4} \alpha=1-2 \sec ^{2} \alpha \) c. \( (\operatorname{sen} \alpha-\cos \alpha)^{2}+(\operatorname{sen} \alpha+\cos \alpha)^{2}=2 \) d. \( \operatorname{ctg}=\operatorname{tg} \alpha=1 \)

Solve the equation by following steps: - step0: Solve for \(\theta\): \(\tan^{4}\left(\theta \right)-\frac{1}{\cos^{4}\left(\theta \right)}=1-\left(2\times \frac{1}{\cos^{2}\left(\theta \right)}\right)\) - step1: Find the domain: \(\tan^{4}\left(\theta \right)-\frac{1}{\cos^{4}\left(\theta \right)}=1-\left(2\times \frac{1}{\cos^{2}\left(\theta \right)}\right),\theta \neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\) - step2: Multiply the terms: \(\tan^{4}\left(\theta \right)-\frac{1}{\cos^{4}\left(\theta \right)}=1-\frac{2}{\cos^{2}\left(\theta \right)}\) - step3: Rewrite the expression: \(\left(\frac{\sin\left(\theta \right)}{\cos\left(\theta \right)}\right)^{4}-\frac{1}{\cos^{4}\left(\theta \right)}=1-\frac{2}{\cos^{2}\left(\theta \right)}\) - step4: Simplify: \(\frac{\sin^{4}\left(\theta \right)}{\cos^{4}\left(\theta \right)}-\frac{1}{\cos^{4}\left(\theta \right)}=1-\frac{2}{\cos^{2}\left(\theta \right)}\) - step5: Multiply both sides of the equation by LCD: \(\left(\frac{\sin^{4}\left(\theta \right)}{\cos^{4}\left(\theta \right)}-\frac{1}{\cos^{4}\left(\theta \right)}\right)\cos^{4}\left(\theta \right)=\left(1-\frac{2}{\cos^{2}\left(\theta \right)}\right)\cos^{4}\left(\theta \right)\) - step6: Simplify the equation: \(\sin^{4}\left(\theta \right)-1=\cos^{2}\left(\theta \right)\cos^{2}\left(\theta \right)-2\cos^{2}\left(\theta \right)\) - step7: Calculate: \(\sin^{4}\left(\theta \right)-1=\cos^{4}\left(\theta \right)-2\cos^{2}\left(\theta \right)\) - step8: Move the expression to the left side: \(\sin^{4}\left(\theta \right)-1-\left(\cos^{4}\left(\theta \right)-2\cos^{2}\left(\theta \right)\right)=0\) - step9: Calculate: \(\sin^{4}\left(\theta \right)-1-\cos^{4}\left(\theta \right)+2\cos^{2}\left(\theta \right)=0\) - step10: Factor the expression: \(\left(-\cos^{2}\left(\theta \right)-\sin^{2}\left(\theta \right)+1\right)\left(\cos^{2}\left(\theta \right)-\sin^{2}\left(\theta \right)-1\right)=0\) - step11: Separate into possible cases: \(\begin{align}&-\cos^{2}\left(\theta \right)-\sin^{2}\left(\theta \right)+1=0\\&\cos^{2}\left(\theta \right)-\sin^{2}\left(\theta \right)-1=0\end{align}\) - step12: Solve the equation: \(\begin{align}&\theta \in \mathbb{R}\\&\theta =k\pi ,k \in \mathbb{Z}\end{align}\) - step13: Find the union: \(\theta \in \mathbb{R}\) - step14: Check if the solution is in the defined range: \(\theta \in \mathbb{R},\theta \neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\) - step15: Find the intersection: \(\theta \neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\) Parece que hubo un problema al intentar resolver las ecuaciones. Vamos a abordar cada una de ellas paso a paso. ### a. \( (1+\csc \alpha) \cdot(1-\sin \alpha)=\cot \alpha \cdot \csc \alpha \) 1. **Reescribimos la ecuación**: \[ (1+\frac{1}{\sin \alpha}) \cdot (1-\sin \alpha) = \frac{\cos \alpha}{\sin \alpha} \cdot \frac{1}{\sin \alpha} \] 2. **Multiplicamos ambos lados**: - Lado izquierdo: \[ (1+\frac{1}{\sin \alpha})(1-\sin \alpha) = 1 - \sin \alpha + \frac{1}{\sin \alpha} - 1 = \frac{1 - \sin^2 \alpha}{\sin \alpha} = \frac{\cos^2 \alpha}{\sin \alpha} \] - Lado derecho: \[ \frac{\cos \alpha}{\sin \alpha} \cdot \frac{1}{\sin \alpha} = \frac{\cos \alpha}{\sin^2 \alpha} \] 3. **Igualamos**: \[ \frac{\cos^2 \alpha}{\sin \alpha} = \frac{\cos \alpha}{\sin^2 \alpha} \] 4. **Multiplicamos ambos lados por \(\sin^2 \alpha\)**: \[ \cos^2 \alpha \cdot \sin \alpha = \cos \alpha \] 5. **Factorizamos**: \[ \cos \alpha (\cos \alpha \sin \alpha - 1) = 0 \] 6. **Soluciones**: - \(\cos \alpha = 0\) o \(\cos \alpha \sin \alpha - 1 = 0\) ### b. \( \tan^4 \alpha - \sec^4 \alpha = 1 - 2 \sec^2 \alpha \) 1. **Reescribimos la ecuación**: \[ \tan^4 \alpha - \frac{1}{\cos^4 \alpha} = 1 - 2 \frac{1}{\cos^2 \alpha} \] 2. **Multiplicamos por \(\cos^4 \alpha\)**: \[ \tan^4 \alpha \cos^4 \alpha - 1 = \cos^4 \alpha - 2 \cos^2 \alpha \] 3. **Sustituimos \(\tan^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha}\)**: \[ \frac{\sin^4 \alpha}{\cos^4 \alpha} - 1 = \cos^4 \alpha - 2 \cos^2 \alpha \] 4. **Simplificamos y resolvemos**. ### c. \( (\sin \alpha - \cos \alpha)^2 + (\sin \alpha + \cos \alpha)^2 = 2 \) 1. **Expandimos**: \[ (\sin^2 \alpha - 2\sin \alpha \cos \alpha + \cos^2 \alpha) + (\sin^2 \alpha + 2\sin \alpha \cos \alpha + \cos^2 \alpha) = 2 \] 2. **Simplificamos**: \[ 2(\sin^2 \alpha + \cos^2 \alpha) = 2 \] 3. **Usamos la identidad**: \[ 2(1) = 2 \] ### d. \( \cot \alpha = \tan \alpha = 1 \) 1. **Esto implica**: \[ \tan \alpha = 1 \implies \alpha = \frac{\pi}{4} + k\pi, k \in \mathbb{Z} \] 2. **Y**: \[ \cot \alpha = 1 \implies \alpha = \frac{\pi}{4} + k\pi, k \in \mathbb{Z} \] ### Resumen de soluciones: - Para la ecuación a, se requiere resolver \(\cos \alpha = 0\) o \(\cos \alpha \sin \alpha - 1 = 0\). - Para la ecuación b, se debe simplificar y resolver. - La ecuación c es siempre verdadera. - Para la ecuación d, \(\alpha = \frac{\pi}{4} + k\pi\). Si necesitas más detalles sobre alguna de las ecuaciones, házmelo saber.