How do you solve this trig identity

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▸ ▸ ▸ ▸ » How do you solve this trig identity

luffy Posts: 3, Reputation: 1

New Member

#1

Apr 10, 2007, 08:11 PM

how do u solve this trig identity

1/sinxcosx-cosx/sinx=tanx

galactus Posts: 2,271, Reputation: 282

Ultra Member

#2

Apr 11, 2007, 06:53 AM

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

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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