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JCF · Apr 9, 2008, 04:03 PM

Hello everyone:
Could you help me verify these identities:

(tan-sin*cos)/sin^2 = tan

(1-sin^2-cos^2+sin^2*cos^3)/sin = sin^2*cos^2*cot

Thanks in advance,
Peter

galactus · Apr 9, 2008, 04:17 PM

#1:

We'll turn the left side into the right side.

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

Replace tan(x) with sin(x)/cos(x):

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

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

Cross-multiply:

$\frac{sin^{2}(x)-sin^{2}(x)cos^{2}(x)}{sin^{3}(x)cos(x)}$

$=\frac{sin^{2}(x)(1-cos^{2}(x))}{sin^{3}(x)cos(x)}$

$=\frac{sin^{4}(x)}{sin^{3}(x)cos(x)}$

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