Question

Prove the trig identity cotx*sec^4(x)=cotx+2tanx+tan^3x (Help, I'm stuck).

Answer 1

I got to
cosx/sinx*1/cos^4x=cosx/sinx+2sinx/cosx+sin^3x/cos^3x
1/(sinx*cos^3x)=cosx/sinx+2sinx/cosx+sin^3x/cos^3x
What should I do next? @wikiemol

Answer 2

oh you are already there! cos(x)/sin(x) = cot(x) and 2sin(x)/cos(x) = 2tanx and sin^3(x)/cos^3(x) = tan^3(x)

Answer 3

What do I do next?

Answer 4

Anyone?

Answer 5

left side:
\[\frac{\cos x}{\sin x}\times\frac{1}{\cos ^4x}=\csc x \sec ^3x\]

Answer 6

Right side:
\[\cot x+\tan x(2+\tan ^2x)=\cot x+\tan x(1+1+\tan ^2x)=\cot x+\tan x(1+\sec ^2x)\]

Answer 7

^ Can you retype that? It got cut off.

Answer 8

\[\cot x+\tan x(1+\sec ^2x)=\]

Answer 9

Oh, I seem to get it...

Answer 10

\[\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}(1+\frac{1}{\cos ^2x})\]

Answer 11

\[\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}+\frac{\sin x}{\cos ^3x}\]

Answer 12

\[\frac{\cos ^2x+\sin ^2x}{\sin x \cos x}+\frac{\sin x}{\cos ^3x}\]

Answer 13

\[\frac{1}{\sin x \cos x}+\frac{\sin x}{\cos ^3x}\]

Answer 14

\[\frac{\cos ^2x}{\sin x \cos ^3x}+\frac{\sin ^2x}{\sin x \cos ^3x}\]

Answer 15

\[\frac{1}{\sin x \cos ^3x}=\csc x \sec ^3x\]

Answer 16

I get it, thank you! @Mertsj

Answer 17

yw