Question

Verify that cot x sec4x = cot x + 2 tan x + tan3x

Answer 1

so i know cotx is cos/sin

Answer 2

idk if we can change the tan though to sin/cos?

Answer 3

this one is hard but try this

Answer 4

first divide everything by cot x

Answer 5

\[\sec^{4} x = 1 + 2\frac{ \tan x }{ \cot x } + \frac{ \tan^{3} x }{ \cot x }\]

Answer 6

and we know that

\[\frac{ \tan x }{ \cot x } = \frac{ \frac{ \sin x }{ \cos x } }{ \frac{ \cos x }{ \sin x } } = \frac{\sin^{2} x }{ \cos^{2} x } = \tan^{2} x\]

Answer 7

would it apply to the second one but it would be (sin^3x/cos^3x)/cosx/sinx ?

Answer 8

that would simplify it to:

\[\sec^{4} x = 1 + 2 \tan^{2} x + \tan^{4} x\]

Answer 9

yeah the second one is:

\[\frac{ \tan^{3} x }{ \cot x } = \frac{ \tan x }{ \cot x } * \tan^{2} x = \tan^{2} x * \tan^{2} x = \tan^{4} x\]

Answer 10

where did you get the second tan^2x

Answer 11

I divided tan^3 x into tan^x * tan^2 x