The trigonometric expression [tex]\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}[/tex] is equivalent to the trigonometric expression [tex]\sec \alpha \cdot \csc \alpha + 1[/tex].
How to prove a trigonometric equivalence
In this problem we must prove that one side of the equality is equal to the expression of the other side, requiring the use of algebraic and trigonometric properties. Now we proceed to present the corresponding procedure:
[tex]\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}[/tex]
[tex]\frac{\tan^{2}\alpha}{\tan \alpha - 1} + \frac{\frac{1}{\tan^{2}\alpha} }{\frac{1}{\tan \alpha} - 1 }[/tex]
[tex]\frac{\tan^{2}\alpha}{\tan \alpha - 1} - \frac{\frac{1 }{\tan \alpha} }{\tan \alpha - 1}[/tex]
[tex]\frac{\frac{\tan^{3}\alpha - 1}{\tan \alpha} }{\tan \alpha - 1}[/tex]
[tex]\frac{\tan^{3}\alpha - 1}{\tan \alpha \cdot (\tan \alpha - 1)}[/tex]
[tex]\frac{(\tan \alpha - 1)\cdot (\tan^{2} \alpha + \tan \alpha + 1)}{\tan \alpha\cdot (\tan \alpha - 1)}[/tex]
[tex]\frac{\tan^{2}\alpha + \tan \alpha + 1}{\tan \alpha}[/tex]
[tex]\tan \alpha + 1 + \cot \alpha[/tex]
[tex]\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} + 1[/tex]
[tex]\frac{\sin^{2}\alpha + \cos^{2}\alpha}{\cos \alpha \cdot \sin \alpha} + 1[/tex]
[tex]\frac{1}{\cos \alpha \cdot \sin \alpha} + 1[/tex]
[tex]\sec \alpha \cdot \csc \alpha + 1[/tex]
The trigonometric expression [tex]\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}[/tex] is equivalent to the trigonometric expression [tex]\sec \alpha \cdot \csc \alpha + 1[/tex].
To learn more on trigonometric expressions: https://brainly.com/question/10083069
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