Question

cotxsec^4x=cotx+2tanx+tan^3x

Answer 1

Solve by trig equations. Make the left equal the right.

Answer 2

you might try converting each term into sin's or cos's

Answer 3

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

Answer 4

Typically, the instructions for this type of problem would be, "Prove the following identity using trig identities."

I would immediately convert cot x into (cos x)/(sin x) and tan x into (sin x)/cos x) for simplicity, although we don't HAVE TO do that.

Here's what your equation would look like now:

\[\frac{ \cos x }{ \sin x }*\frac{ 1 }{ \cos ^{4}x }=\frac{ \cos x }{ \sin x }+2\frac{ \sin x }{ \cos x }+\frac{ \sin ^{3}x }{\cos ^{3}x .}\]

I'd suggest you try factoring (cos x) / (sin x) out of the right side, just as it has been factored out of the left side.

Answer 5

\[\cot x(1+2\tan ^{2}x+\tan ^{4}x)\]

Answer 6

\[\cot x=\frac{ 1 }{ \tan x }\]

Answer 7

\[\cot x(1+2\tan ^{2}x+\tan ^{4}x)=(\cot x+2\tan ^{2}x \cot x+\cot x \tan ^{4}x)\]=\[\cot x+2 \tan x+\tan^3x\]

Answer 8

or if u write s = sin x and c = cos x ( because i;m lazy)

referring to mathmales post:
left hand side = c / s * c^4 = 1 / s * c^3

right hand side = c/s + 2s/c + s^3/c3
convert to single fraction gives
c^4 + 2s c^2 + s^4
------------------
s * c^3

= (c^2 + s^2)(c^2 + s^2)
------------------- = 1 / s * c*3 (as s^2 + c^2 = 1
s * c^3


so LHS = RHS