how do I simplify these?

Question

1. (Cot/1-tan) + (Tan/1-Cot) - Tan - Cot

2. (1+cos) (csc-cot)

Answer 1

sloppy notation.

the cot of what, the sin of what?
sin, cos, tan, etc are mathematical operators
I will use x as the "angle"

(Cotx/1-tanx) + (Tanx/1-Cotx) - Tanx - Cotx

I usually change everybody to sines and cosines, so ...

= (cosx/sinx)/(1 - sinx/cosx) + (sinx/cosx)/(1 - cosx/sinx) - sinx/cox - cosx/sinx
= (cos^2 x)/(sinx(cosx - sinx)) + (sin^2x)/(cosx(sinx-cosx) - sinx/cosx - cosx/sinx

= (cos^2 x)/(sinx(cosx - sinx)) - (sin^2x)/(cosx(cosx - sinx) - sinx/cosx - cosx/sinx

= .lots of messy typing here

form a common denominator of
(sinx)(cosx)(cosx-sinx) and try to finish it.

Answer 2

for the second, I would use the same approach.

(1+cosx) (cscx-cotx)
= (1 + cosx)(1/sinx - cosx/sinx)
= 1/sinx - cosx/sinx + cosx/sinx - cos^2x/sinx
= 1/sinx(1 - cos^2x)
= 1/sinx(sin^2x_
= sinx

Answer 3

(cos/sin/(1 - sin/cos) + sin/cos/(1-cos/sin) - sin/cos - cos /sin
multiply top and bottom of all by sin cos
cos^2/(sin cos -sin^2) + sin^2/(sin cos - cos^2) - (sin^2+cos^2)/sin cos

cos^2/(sin cos -sin^2) + sin^2/(sin cos - cos^2) - 1 /sin cos

cos^2/(sin (cos -sin)) - sin^2/(cos(cos - sin)) - 1 /sin cos

writing cos as c and sin as s
just doing first two terms for now
c^2/(s(c-s)) - s^2/(c(c-s))

1/(c-s) * (c^2/s -s^2/c)
1/(c-s) * ( (c^3-s^3)/sc)
1/(c-s) * ((c-s)(c^2 + sc + s^2)/sc
(1+sc)/sc
now put that -1/sc back
(1+sc)/sc - 1/sc
sc/sc
1 Caramba !!!!

Answer 4

It is the sins, cos, etc of theta

Answer 5

OK, I was not about to type theta all the time either, and in fact even got tired of typing sin and cos.