What is the answer to this Trig. problem?

Question

(tan x + cot x)/sec² x . I keep getting cos² x for the answer, but whenever I plug in an angle, I get two different answers. I multiply both sides by cos²x (inverse of sec²x), and then do a process of elimination with the cosines and sines, and get cos²x. What is the correct answer to this problem?

Answer 1

Bottom line sec2x = 1/cos2x

therefore equation can be rewritten as:

(tan x + cot x)* cos2x
= (sinx/cosx + cosx/sinx)*cos2x
= sinxcosx + cos3x/sinx
=(sin2xcosx+cos3x)/sinx
=(cosx(sin2x+cos2x))/sinx
=cosx/sinx
= cotx

Answer 2

TAN = SIN/COS AND COT = COS/SIN

AND SEC^2 = 1/COS^2,

I seem to remember the above,

dont think when you plug them in you are going to get Cos^2X.
I cant follow the guy above.

But I did play with the cos and sin, and yes the answer becomes Cotx.

you need to put the entire equation under a commom demonator. that that will be sin *cos , and ofcourse mult both sides by Cos^2x.

Answer 3

(sinx/cosx+cosx/sinx)*cos^2x = (sin^2+cos^2) *cos/sin=cotanx