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

#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)