Respuesta :

arthurpdc
Manipulating the left side of the equation, we obtain:

[tex]\dfrac{1}{\sin x}-\dfrac{1}{\csc x}=\dfrac{\csc x-\sin x}{\sin x\cdot\csc x} [/tex]

Using that [tex]\csc x=\dfrac{1}{\sin x}[/tex]:

[tex]\dfrac{\csc x-\sin x}{\sin x\cdot\csc x}=\dfrac{\csc x-\sin x}{\sin x\cdot\dfrac{1}{\sin x}}=\dfrac{\csc x-\sin x}{1}=\csc x-\sin x\\\\\boxed{\dfrac{1}{\sin x}-\dfrac{1}{\csc x}=\csc x-\sin x}~~\blacksquare[/tex]
[tex]\mathrm{csc}\,x = \frac{1}{\sin x}[/tex], so:

[tex]\dfrac{1}{\sin x}-\dfrac{1}{\mathrm{csc}\,x} = \dfrac{1}{\sin x} - \frac{1}{\frac{1}{\sin x}} = \dfrac{1}{\sin x} - \sin x = \mathrm{csc}\,x -\sin x \quad \square[/tex]

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