Question

2 2
Prove that
tan
| - coto
+ coto
= 1 sec o coseca​

Answer 1

Answer:

[math]\dfrac{\tan A}{1-\cot A} + \dfrac{\cot A}{1-\tan A} = \dfrac{\frac{\sin A}{\cos A}}{1-\frac{\cos A}{\sin A}} + \dfrac{\frac{\cos A}{\sin A}}{1-\frac{\sin A}{\cos A}} [/math]

[math]= \dfrac{{\sin}^2 A}{\cos A(\sin A-\cos A)} + \dfrac{{\cos}^2 A}{\sin A(\cos A-\sin A)} [/math]

[math]= \dfrac{{\sin}^3 A-{\cos}^3 A}{\cos A \sin A(\sin A-\cos A)} [/math]

[math]= \dfrac{{\sin}^2 A+\cos A \sin A+{\cos}^2 A}{\cos A \sin A} [/math]

[math]= \dfrac{1+\cos A \sin A}{\cos A \sin A} [/math]

[math]= 1+\sec A \csc A. \blacksquare [/math]

Step-by-step explanation:

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