Ask Me Help Desk - View Single Post - (cot x/1-tan x) +(tan x/1-cot x) = csc x sec x+1

galactus · Dec 11, 2010, 11:57 AM

$\frac{cot(x)}{1-tan(x)}+\frac{tan(x)}{1-cot(x)}=csc(x)sec(x)+1$

Transform everything into sin and cos. This is a good idea when solving most any trig identity.

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

$\frac{\frac{sin(x)}{cos(x)}}{\frac{sin(x)-cos(x)}{sin(x)}}$ $+\frac{\frac{cos(x)}{sin(x)}}{\frac{cos(x)-sin(x)}{cos(x)}}$

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

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

Note the difference of two cubes in the numerator. Factoring and cancelling leads to:

$\frac{sin^{2}(x)+sin(x)cos(x)+cos^{2}(x)}{sin(x)co s(x)}$

$\frac{1+sin(x)cos(x)}{sin(x)cos(x)}$

$\frac{1}{sin(x)cos(x)}+\frac{sin(x)cos(x)}{sin(x)c os(x)}$

$sec(x)csc(x)+1$