Trigo Identity, I Need Help ASAP?

Question

I need help on this problem, I would be so thankful if you could show me the steps to the final answer.

Work one side of the equation:

(sec x/cosx) - (tan x/cot x) = 1

Answer 1

sec x = 1 / cos x
tan x = sin x / cos x
cot x = 1 / tan x = cos x / tan x

So

(sec x / cos x) - (tan x / cot x)
= ((1 / cos x) / cos x) - ( (sin x / cos x) / (cos x / sin x))
= (1 / (cos x)^2) - ((sin x)^2 / (cos x)^2)
= (1 - (sin x)^2) / (cos x)^2
= (cos x)^2 / (cos x)^2
= 1

Answer 2

Here are some things to remember when solving this type of problem:
1) work the hardest side first (in this case the left side)
2) change all terms to cosx and sinx

secx = 1/cosx
tanx = sinx/cosx
cotx = cosx/sinx
sin^2x + cos^2x = 1, keep in mind: 1-sin^2x = cos^2x

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

hope i helped.

Answer 3

1. sec/cos - tan/cot = (1/cos)/cos - (sin/cos)/(cos/sin)
= 1/(cos^2) - sin^2/(cos^2)
= (1-sin^2)/ (cos^2)
= (cos^2) / cos^2
=1

Answer 4

the 1st id is plenty off. cos(a + b) = cos(a)cos(b) - sin(a)sin(b) This follows from the sin(a + b) = sin(a)cos(b) + cos(a)sin(b), on the id That cos(x) = sin(Pi/2 - x), plug it in into that formula to get the above style. 2. you place the equations in an unsightly style which will cause them to slightly harder to tutor. to restoration this, substitute t = x/2. (no longer that we lose no generality in view that the two x/2 and x run the full genuine numbers set). sin(2t) = 2sin(t)cos(t) initiate off with the sin(a + b) formula, in basic terms this difficulty it particularly is going to be 2t = t + t sin(t + t) = sin(t)cos(t) + cos(t)sin(t) = 2sin(t)cos(t) substitute back to get your style, sin(x) = 2sin(x/2)cos(x/2) For cos(2t), we will use the cos(a + b) formula with t + t. cos(t + t) = cos(t)cos(t) - sin(t)sin(t) = cos^(t) - sin^(t) substitute back, cos(x) = cos^2(x/2) - sin^2(x/2) Q.E.D.