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how do u solve this trig identity

1/sinxcosx-cosx/sinx=tanx

$\frac{1}{sin(x)cos(x)}-\frac{cos(x)}{sin(x)}=tan(x)$

Cross multiply left side:

$\frac{sin(x)-sin(x)cos^{2}(x)}{sin^{2}(x)cos(x)}$

Factor out sinx:

$\frac{\sout{sin(x)}(\overbrace{1-cos^{2}(x)}^{\text{sin^2(x)}})}{\sout{sin(x)}(sin( x)cos(x)}$

$\frac{sin^{2}(x)}{sin(x)cos(x)}=\frac{sin(x)}{cos( x)}=tan(x)$

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