Question

Prove identity trigonometric equation

Answer 1

Explanation:

The given equation is False, so cannot be proven to be true.

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Perhaps you want to prove ...

  $2\tan{x}=\dfrac{\cos{x}}{\csc{(x)}-1}+\dfrac{\cos{x}}{\csc{(x)}+1}$

This is one way to show it:

  $2\tan{x}=\cos{(x)}\dfrac{(\csc{(x)}+1)+(\csc{(x)}-1)}{(\csc{(x)}-1)(\csc{(x)}+1)}\\\\=\cos{(x)}\dfrac{2\csc{(x)}}{\csc{(x)}^2-1}=2\cos{(x)}\dfrac{\csc{x}}{\cot{(x)}^2}=2\dfrac{\cos{(x)}\sin{(x)}^2}{\cos{(x)}^2\sin{(x)}}\\\\=2\dfrac{\sin{x}}{\cos{x}}\\\\2\tan{x}=2\tan{x}\qquad\text{QED}$

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We have used the identities ...

  csc = 1/sin

  cot = cos/sin

  csc^2 -1 = cot^2

  tan = sin/cos