help with trigonometry adition?

Question

1/cos^2X - 1/cot^2X = ?
it's either: 1-sinX, 1, or -sec^2X

2tanX - (1+tanX)^2 = ?
it's either 0, 1, or -sec^2X

tan X (cotX - cosX) = ?
it's either 0, 1-sinX, or 1

Answer 1

NOTE:

1 / cos = sec

1 / cot = tan

and

sec^2 - tan^2 = 1



(1 / cos^2 x) - 1 / (cot^2 x)

= sec^2 x - tan^2 x

= 1

**********************

NOTE:

sec^2 - tan^2 = 1





2 tan x - (1 + tan x)^2

= 2 tan x - (1 + 2 tan x + tan^2 x)

= 2 tan x - 1 - 2 tan x - tan^2 x

= -1 - tan^2 x

= (-1)(1 + tan^2 x)

= (-1)(sec^2 x)

= -sec^2 x



***************

tan x = 1 / cot x

and

tan x = sin x / cos x





(tan x)(cot x - cos x)

= (tan x)(cot x) - (tan x)(cos x)

= 1 - (sin x / cos x)(cos x)

= 1 - sin x

Answer 2

cot x = cos x/ sinx

so



1/cos^2x - 1/cos^2x/sin^2x



1/cos^2x - sin^2x/cos^2x



1-sin^2x /cos^2x



but:



sin^2x + cos^2x = 1 and so



1-sin^2x = cos^2x therefore substituting cos^2x for 1-sin^2x you get



cos^2x/cos^2x = 1



***



2tanx - (1+tanx)^2 =



2tanx - [1 + 2tanx + tan^2x] =



2tanx - 1 - 2tanx - tan^2x



the 2tanx cancel so you're left with



-1-tan^2x = -(1+tan^2x) but 1+tan^2x = sec^2x so:



-sec^2x

Answer 3

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

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



2tan(x)-(1+tan^2(x)+2tan(x))

=2tan(x)-1-tan^2(x)-2tan(x)

=-1-tan^2(x)

=-(1+tan^2(x))

=-sec^2(x)