Question

Could someone assist me with this Trigonometric equation? I would have to make one side equal to the other Vice Versa

cot x sec^4x = cot x + 2 tan x + tan^3x

After I chosen to replace sec^4x with (1+tan^2x)^2 then (1+tanx)(1+tanx) . After foiling I'm left with (2 tanx + tan x^2). Now I have this cot left over. What exactly do I do with this? If there is any mistakes in my reasoning please correct them.

Answer 1

You wrote:(1+tan x)(1+tan x). \[(1+\tan x)(1+\tan x)=1+2\tan x + \tan^2 x\]Where would you put this in?

I would reason as follows: replace cotx by cosx/sinx and secx by 1/cosx:\[\frac{ \cos x }{ \sin x }\cdot \frac{ 1 }{ \cos^4x }=\frac{ \cos x }{ \sin x }+\frac{ 2\sin x }{ \cos x }+\frac{ \sin^3x }{ \cos^3x }\]The left hand side is equal to\[\frac{ 1 }{ \sin x \cos^3x }\]The right hand side consists of three fractions. If you make the denominators the same and add, you also get a fraction with denominator sin x cos³x, so there's some light there...

Answer 2

nice work@tushara

Answer 3

nice cam @Tushara

Answer 4

thnx guys, i hope u understand every step toadsage

Answer 5

Looking at both of your examples I see that you both have 1/sin x cos^3 x at hand. But looking at Tsuhara's picture I don't understand how you got from 1 / sin x cos^ 3 to cos x / sin x cos 4 x. Nice picture though !

Answer 6

Multiply numerator and denominator with cos x

Answer 7

Great, I see that now. If it's not too much to ask, why exactly do you have to multiply it by cos x?

Answer 8

simplify (cotx)(sec^4x) till u get to 1/(sinx cos^3x) so wen u work on the RHS of the equation, after u get to 1/(sinx cos^3x) work backwards on each step u took earlier

Answer 9

Thanks a lot everyone!

Answer 10

yw!