(sec^2x-tanx)/1-cosx?

Question

(sec^2x-tanx)/1-cosx

in case anyone is confused, its secant squaredx - tangent x all over 1 - cosine x

i need to simplify using trig identities (pythagorean identies and basic stuff)

how do i do it, completely confused.

the answer can be unsolvable btw

Answer 1

Sorry, the answer of mlam18 is false, because sinx/cosx is different of sin²x/cos²x.

But I don't find a good solution !

1/cos²x - tanx = 1/cos²x - sinx/cosx
= (1 - sinx cosx) / cos²x

then
(sec²x-tanx) / (1 - cosx)
= (1 - sinx cosx) / [cos²x(1 - cosx)]
This expression seems impossible to simplify,
I had tried many solutions (with sin(x/2) ans cos(x/2), or sin(2x) and cos(2x)) and I had not found a good answer.

I will try another way ! Perhaps !?

Answer 2

(sec^2x-tanx)/1-cosx ------------(1)

sec²x - tan x = 1/cos²x - sin x/cos x
= 1/cos²x - sin²x/ cos²x
= 1- sin²x / cos²x
= cos²x /cos²x
= 1

(1) becomes 1/ 1-cos x