Question

Proving trig equations (again)

\(\sf cotx~sec^4x=cotx+2tanx+tan^3x\)

Answer 1

@ganeshie8

Answer 2

lol it was faulty

\(\large\tt \begin{align} \color{black}{
= \cot x\sec^4x\\~\\
=\cot x(\sec^2x)^2\\~\\
=\cot x(1+\tan^2x)^2\\~\\
=\cot x(1+2\tan^2x+\tan^4x)\\~\\
}\end{align}\)

Answer 3

If I multiply both sides by tan x, I get:
\[\frac{ 1 }{ (\cos x)^{4} }=1+2(\tan x)^{2}+(\tan x)^{4}=[1+(\tan x )^{2}]^{2}\]

Answer 4

Yeah, there we go @mathmath333

Answer 5

yea its ok , now i think

Answer 6

I see where that is coming from, but how would that become
\(\sf cotx+2tanx+tan^3x\)?

Answer 7

further u have to multiply \(\cot x\) in the bracket

and use this \(\cot x=\dfrac{1}{\tan x}\)