\( \cos ^ { 2 } \alpha + ( \frac { \cos \alpha } { \tan \alpha } ) ^ { 2 } = \frac { 1 } { \tan ^ { 2 } a } \)

Solve the equation by following steps: - step0: Solve for \(\alpha\): \(\cos^{2}\left(\alpha \right)+\left(\frac{\cos\left(\alpha \right)}{\tan\left(\alpha \right)}\right)^{2}=\frac{1}{\tan^{2}\left(\alpha \right)}\) - step1: Find the domain: \(\cos^{2}\left(\alpha \right)+\left(\frac{\cos\left(\alpha \right)}{\tan\left(\alpha \right)}\right)^{2}=\frac{1}{\tan^{2}\left(\alpha \right)},\alpha \neq \frac{k\pi }{2},k \in \mathbb{Z}\) - step2: Rewrite the expression: \(\cos^{2}\left(\alpha \right)+\frac{\cos^{2}\left(\alpha \right)}{\tan^{2}\left(\alpha \right)}=\frac{1}{\tan^{2}\left(\alpha \right)}\) - step3: Rewrite the expression: \(\cos^{2}\left(\alpha \right)+\frac{\cos^{2}\left(\alpha \right)}{\left(\frac{\sin\left(\alpha \right)}{\cos\left(\alpha \right)}\right)^{2}}=\frac{1}{\left(\frac{\sin\left(\alpha \right)}{\cos\left(\alpha \right)}\right)^{2}}\) - step4: Simplify: \(\cos^{2}\left(\alpha \right)+\frac{\cos^{4}\left(\alpha \right)}{\sin^{2}\left(\alpha \right)}=\frac{1}{\left(\frac{\sin\left(\alpha \right)}{\cos\left(\alpha \right)}\right)^{2}}\) - step5: Expand the expression: \(\cos^{2}\left(\alpha \right)+\frac{\cos^{4}\left(\alpha \right)}{\sin^{2}\left(\alpha \right)}=\frac{1}{\frac{\sin^{2}\left(\alpha \right)}{\cos^{2}\left(\alpha \right)}}\) - step6: Calculate: \(\cos^{2}\left(\alpha \right)+\frac{\cos^{4}\left(\alpha \right)}{\sin^{2}\left(\alpha \right)}=\frac{\cos^{2}\left(\alpha \right)}{\sin^{2}\left(\alpha \right)}\) - step7: Multiply both sides of the equation by LCD: \(\left(\cos^{2}\left(\alpha \right)+\frac{\cos^{4}\left(\alpha \right)}{\sin^{2}\left(\alpha \right)}\right)\sin^{2}\left(\alpha \right)=\frac{\cos^{2}\left(\alpha \right)}{\sin^{2}\left(\alpha \right)}\times \sin^{2}\left(\alpha \right)\) - step8: Simplify the equation: \(\cos^{2}\left(\alpha \right)\sin^{2}\left(\alpha \right)+\cos^{4}\left(\alpha \right)=\cos^{2}\left(\alpha \right)\) - step9: Rewrite the expression: \(\cos^{2}\left(\alpha \right)=\cos^{2}\left(\alpha \right)\) - step10: Cancel equal terms: \(0=0\) - step11: The statement is true: \(\alpha \in \mathbb{R}\) - step12: Check if the solution is in the defined range: \(\alpha \in \mathbb{R},\alpha \neq \frac{k\pi }{2},k \in \mathbb{Z}\) - step13: Find the intersection: \(\alpha \neq \frac{k\pi }{2},k \in \mathbb{Z}\) To solve the equation \[ \cos ^ { 2 } \alpha + \left( \frac { \cos \alpha } { \tan \alpha } \right) ^ { 2 } = \frac { 1 } { \tan ^ { 2 } \alpha }, \] we can start by simplifying the left-hand side. 1. **Understanding the terms**: - Recall that \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\). - Therefore, \(\tan^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha}\). 2. **Substituting \(\tan \alpha\)**: - The term \(\frac{1}{\tan^2 \alpha}\) can be rewritten as \(\frac{\cos^2 \alpha}{\sin^2 \alpha}\). 3. **Rewriting the equation**: - The left-hand side becomes: \[ \cos^2 \alpha + \left( \frac{\cos \alpha}{\frac{\sin \alpha}{\cos \alpha}} \right)^2 = \cos^2 \alpha + \left( \frac{\cos^2 \alpha}{\sin \alpha} \right)^2. \] 4. **Combining terms**: - The equation simplifies to: \[ \cos^2 \alpha + \frac{\cos^4 \alpha}{\sin^2 \alpha} = \frac{\cos^2 \alpha}{\sin^2 \alpha}. \] 5. **Setting the equation**: - Rearranging gives: \[ \cos^2 \alpha + \frac{\cos^4 \alpha}{\sin^2 \alpha} - \frac{\cos^2 \alpha}{\sin^2 \alpha} = 0. \] 6. **Factoring out common terms**: - This leads to: \[ \cos^2 \alpha \left( 1 + \frac{\cos^2 \alpha - 1}{\sin^2 \alpha} \right) = 0. \] 7. **Finding solutions**: - The first factor gives \(\cos^2 \alpha = 0\), which implies \(\alpha = \frac{\pi}{2} + k\pi\) for \(k \in \mathbb{Z}\). - The second factor leads to a contradiction since \(1 + \frac{\cos^2 \alpha - 1}{\sin^2 \alpha} = 0\) does not yield valid solutions. 8. **Conclusion**: - The only valid solutions are when \(\alpha \neq \frac{k\pi}{2}\) for \(k \in \mathbb{Z}\). Thus, the solution to the equation is: \[ \alpha \neq \frac{k\pi }{2}, k \in \mathbb{Z}. \]