anonymous
  • anonymous
Verify the equation by substituting identities to match the right side to the left side. cot x sec4x = cot x + 2 tan x + tan3x
Mathematics
schrodinger
  • schrodinger
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SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \cot x \sec^4x = \cot x + 2 \tan x + \tan^3x }\) ?
SolomonZelman
  • SolomonZelman
Or, \(\large\color{black}{ \displaystyle \cot x ~\sec(4x) = \cot x + 2 \tan x + \tan(3x ) }\) ?
anonymous
  • anonymous
The first one. :)

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SolomonZelman
  • SolomonZelman
cool.
anonymous
  • anonymous
So I'm trying to do the right side.
anonymous
  • anonymous
And first I changed cot x to 1/tan x, and then cos x/sin x
anonymous
  • anonymous
HI
anonymous
  • anonymous
I'll try to write what I did first. :)
anonymous
  • anonymous
RIGHT SIDE: \[\frac{ \cos x }{ \sin x }+\frac{ 2 \sin x }{ \cos x }+\frac{ \sin ^{3}x }{ \cos ^{3}x }\]
anonymous
  • anonymous
And then I'm kinda stuck with it.
anonymous
  • anonymous
I used the reciprocal and quotient identities by the way.
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \cot x \sec^4x = \cot x + 2 \tan x + \tan^3x }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \cos x }{ \sin x }+\frac{ 2 \sin x }{ \cos x }+\frac{ \sin ^{3}x }{ \cos ^{3}x } }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \cos^4x }{ \sin x \cos^3x}+\frac{ 2 \cos^2x\sin^2x }{ \cos^3x\sin x }+\frac{ \sin ^{4}x }{ \cos ^{3}x \sin x} }\)
anonymous
  • anonymous
I may sound dumb to you, but can you explain how you get that?
SolomonZelman
  • SolomonZelman
then combine the fractions and use (a+b)²=a²+2ab+b²
SolomonZelman
  • SolomonZelman
I just found the common denominator to each fraction.
anonymous
  • anonymous
Ohhh. Thank you!
anonymous
  • anonymous
I will work on it right now. So I will close this question. I'll just ask another question if I'm stuck with something. :)
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \cos^4x }{ \sin x \cos^3x}+\frac{ 2 \cos^2x\sin^2x }{ \cos^3x\sin x }+\frac{ \sin ^{4}x }{ \cos ^{3}x \sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \cos^4x + 2 \cos^2x\sin^2x + \sin ^{4}x }{ \cos ^{3}x \sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \left(\cos^2x+\sin^2x\right)^2 }{ \cos ^{3}x \sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \left(1\right)^2 }{ \cos ^{3}x \sin x} }\)
anonymous
  • anonymous
Please don't tell me the answer. :( I'm trying to know how to do it haha.
SolomonZelman
  • SolomonZelman
oh, sorry, go ahead.
anonymous
  • anonymous
\[\cot x \sec ^{4}x=\frac{ 1 }{ \cos ^{3}xsin x }\] @SolomonZelman What's next?
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x \sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x }\cdot\frac{1}{\sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{4}x }\cdot\frac{\cos x}{\sin x} }\)
SolomonZelman
  • SolomonZelman
then the rest is even easier...
anonymous
  • anonymous
Why did it become like that?
anonymous
  • anonymous
The third one I mean.
SolomonZelman
  • SolomonZelman
i multiplyied by cos(x)/cos(x)
anonymous
  • anonymous
Why? Sorry if I keep on bugging you. Because what I did was I turned 1/sin x to csc x.
SolomonZelman
  • SolomonZelman
oh, don't .... and it's ok
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x \sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x }\cdot\frac{1}{\sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x }\cdot\frac{1}{\sin x} \cdot \frac{\cos x}{\cos x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x\cdot \cos x }\cdot\frac{1\cdot \cos x }{\sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{4}x }\cdot\frac{\cos x}{\sin x} }\)
anonymous
  • anonymous
And from the third one, it will become the equation on the left side, right?
SolomonZelman
  • SolomonZelman
yes, because: 1/cos\(^4\)x = ? cos(x)\(/\)sin(x) = ?
anonymous
  • anonymous
sec^4x and cot x :) Thank you so much for guiding me! I can now do the other ones.
SolomonZelman
  • SolomonZelman
Yeah.... the main trick was in finding the common denominator and noticing a²+2ab+b²=(a+b)² you welcome, and I got a prayer to attend right now. cu

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