Question

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

Answer 1

I changed the sec^4x

Answer 2

i did (sec^2x)^2 which is (1+tan^2x)(1+tan^2x)

Answer 3

and then 1+2tan^2x+tan^4x

Answer 4

and then i got stuck

Answer 5

lets also try right hand side

Answer 6

ok what do you mean?

Answer 7

\[\cot x+2\tan x+\tan 3x=???\]

Answer 8

well the only thing you can do is change cot to cos/sin

Answer 9

consider c as cos and s as sin tell me if its not clear \[\frac{ c }{ s }+\frac{ 2s }{ c }=\frac{ c^2+2s^2 }{ sc }=\frac{ 1-s^2+2s^2 }{ sc }=\frac{ 1-s^2 }{sc }\]

Answer 10

tan x=sinx /cosx cot x=cos x/sin x so I added them up

Answer 11

what about tan^3x

Answer 12

sin 3x /cos 3x

Answer 13

ok so then 1-s^2/sc = cos^2/sinxcosx=cosx/sinx=cot?

Answer 14

wait its 1+s^2/sc

Answer 15

oh

Answer 16

now i'm lost

Answer 17

lets start over I also lost myself there

Answer 18

lol ok

Answer 19

\[\cot x \sec 4x=\cot x(1+\tan 4x)\]
\[\cot x+\cot x \tan 4x=\cot x +\cot x(\frac{ \sin 4x }{ \cos 4x })\]

Answer 20

you are missing the 2tanx

Answer 21

2tan^2x

Answer 22

\[\frac{\sin 4x}{\cos 4x}=\frac{2\cos2x \sin 2x}{2\cos^22x-1}\]

Answer 23

above I started with LHS left hand side so there is no tan initilly

Answer 24

i messed up i give up sry

Answer 25

ok.....thanks