Respuesta :

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Explanation:

Apparently, you're to show this is an identity.

  [tex]\dfrac{\tan^2A}{\tan A-1}+\dfrac{\cot A}{1-\tan A}=1+\sec A\csc A\\\\\dfrac{\tan^2 A-\dfrac{1}{\tan A}}{\tan A-1}=\\\\\dfrac{\tan^3 A-1}{(\tan A)(\tan A-1)}=\\\\\\\dfrac{\tan^2 A+\tan A+1}{\tan A}=\\\\1+\dfrac{\sec^2 A}{\tan A}=\\\\1+\dfrac{1}{\cos A}\cdot\dfrac{1}{\sin A}=1+\sec A\csc A[/tex]

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